NCERT Solutions Class 11 Maths Chapter 10

Q.1 Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Ans

$\begin{array}{l}\mathrm{The}\text{given points are:}\\ \text{A}(-4,5),\mathrm{B}(0,7),\mathrm{C}(5,-5)\text{and D}(-4,-2).\\ \mathrm{Area}\text{of}\mathrm{\Delta ABC}=\left|\frac{1}{2}\{{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\}\right|\\ =\left|\frac{1}{2}\{-4(7+5)+0(-5+5)+5(5-7)\}\right|\\ =\left|\frac{1}{2}(-48+0-10)\right|\\ =29\mathrm{sq}.\text{units}\end{array}$ $\begin{array}{l}\mathrm{Area}\text{of}\mathrm{\Delta ADC}=\left|\frac{1}{2}\{-4(-2+5)-4(-5-5)+5(5+2)\}\right|\\ =\left|\frac{1}{2}(-12+40+35)\right|\\ =\frac{63}{2}\mathrm{sq}.\text{units}\\ \text{\hspace{0.17em}Area}\left(\mathrm{ABCD}\right)=\mathrm{Area}\text{of}\mathrm{\Delta ABC}+\mathrm{Area}\text{of}\mathrm{\Delta ADC}\\ =(29+\frac{63}{2})\mathrm{sq}.\text{units}\\ =\frac{121}{2}\mathrm{sq}.\text{units}\\ \text{Thus, the area of quadrilateral is}\frac{121}{2}\mathrm{square}\text{units.}\end{array}$

Q.2 The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Ans

$\begin{array}{l}\text{Coordinate of B and C are}(0,–\text{a})\text{and}\left(\text{0,a}\right)\mathrm{}\text{respectively}.\\ \mathrm{In}\mathrm{\Delta AOB},\angle \mathrm{AOB}=90\mathrm{°}\\ \mathrm{So},\text{by Pythagoras theorem}\\ {\text{AB}}^{\text{2}}={\mathrm{AO}}^2+{\mathrm{OB}}^2\\ \Rightarrow {\left(2\mathrm{a}\right)}^2={\mathrm{AO}}^2+{\mathrm{a}}^2\\ \Rightarrow \mathrm{AO}=\sqrt{4{\mathrm{a}}^2-{\mathrm{a}}^2}\\ =\pm \sqrt{3}\mathrm{a}\\ \mathrm{So},\text{the coordinate of A is}(0,\sqrt{3}\mathrm{a})\text{or}(0,-\sqrt{3}\mathrm{a}).\\ \mathrm{Therefore},\text{the coordinates of triangle are:}\\ (0,\mathrm{a}),(0,-\mathrm{a})\mathrm{and}(0,\sqrt{3}\mathrm{a})\mathrm{Or}(0,\mathrm{a}),(0,-\mathrm{a})\mathrm{and}(0,-\sqrt{3}\mathrm{a})\end{array}$

Q.3 Find the distance between P (x₁, y₁) and Q (x₂, y₂) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Ans

$\begin{array}{l}\text{Distance between two points P}({\mathrm{x}}_1,{\mathrm{y}}_1)\text{and Q}({\mathrm{x}}_2,{\mathrm{y}}_2)\\ =\left|\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}\right|\\ \left(\mathrm{i}\right){\text{When PQ is parallel to y-axis: Then, x}}_{\text{1}}={\text{x}}_{\text{2}}\\ \text{PQ}=\left|\sqrt{{({\mathrm{x}}_1-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}\right|\\ \text{PQ}=\left|\sqrt{{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}\right|\\ \text{PQ}=\left|({\mathrm{y}}_2-{\mathrm{y}}_1)\right|\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right){\text{When PQ is parallel to x-axis: Then, y}}_{\text{1}}={\text{y}}_{\text{2}}\\ \text{PQ}=\left|\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}\right|\\ \text{PQ}=\left|\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2}\right|\\ \text{PQ}=\left|({\mathrm{x}}_2-{\mathrm{x}}_1)\right|\end{array}$

Q.4 Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Ans

$\begin{array}{l}\mathrm{Let}\text{the point on x-axis be P}(\mathrm{a},0).\\ \mathrm{Given}\text{points are A}(7,6)\text{and B}(3,4).\\ \mathrm{Then},\text{according to given condition:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PA}=\text{PB}\\ \Rightarrow \left|\sqrt{{(\mathrm{a}-7)}^2+{(0-6)}^2}\right|=\left|\sqrt{{(\mathrm{a}-3)}^2+{(0-4)}^2}\right|\\ \Rightarrow \left|\sqrt{{(\mathrm{a}-7)}^2+36}\right|=\left|\sqrt{{(\mathrm{a}-3)}^2+16}\right|\\ \Rightarrow {\mathrm{a}}^2-14\mathrm{a}+49+36={\mathrm{a}}^2-6\mathrm{a}+9+16\\ \Rightarrow 49+36-25=14\mathrm{a}-6\mathrm{a}\\ \Rightarrow 49+36-25=8\mathrm{a}\\ \Rightarrow 60=8\mathrm{a}\\ \Rightarrow \mathrm{a}=\frac{60}{8}\\ \Rightarrow \mathrm{a}=\frac{15}{2}\end{array}$ $Therefore,\text{the point on x-axis is}\left(\frac{15}{2},0\right).$

Q.5 Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Ans

$\begin{array}{l}\mathrm{Mid}-\mathrm{point}\text{of P}(0,-\text{\hspace{0.17em}4})\text{and B}(\text{8},0)=(\frac{0+8}{2},\frac{-4+0}{2})\\ =(4,-2)\\ \mathrm{Slope}\text{of line passing through origin}(0,0)\text{and}(4,-2)\\ =\frac{-2-0}{4-0}\\ =\frac{-2}{4}\\ =-\frac{1}{2}\\ \mathrm{Therefore},\text{the required slope of line is}-\frac{1}{2}.\end{array}$

Q.6 Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Ans

$\begin{array}{l}\text{Given points are:}(\text{4},\text{4}),(\text{3},\text{5})\text{and}(–\text{1},–\text{1})\\ \mathrm{Slope}\text{of}(\text{4},\text{4})\mathrm{and}(\text{3},\text{5}),{\mathrm{m}}_1=\frac{5-4}{3-4}\\ =\frac{1}{-1}\end{array}$ $\begin{array}{l}=-1\\ \mathrm{Slope}\text{of}(\text{3},\text{5})\mathrm{and}(–\text{1},–\text{1}),{\mathrm{m}}_2=\frac{-1-4}{-1-3}\\ =\frac{-5}{-4}\\ =\frac{5}{4}\\ \mathrm{Slope}\text{of}(–\text{1},–\text{1})\mathrm{and}(\text{4},\text{4}),{\mathrm{m}}_3=\frac{4+1}{4+1}\\ =\frac{5}{5}\\ =1\\ \mathrm{Here},{\mathrm{m}}_1.{\mathrm{m}}_3=-1\times 1\\ =-1\\ \mathrm{So},{\text{the sides of triangle having slope m}}_{\text{1}}{\text{and m}}_{\text{3}}\text{are}\\ \text{perpendicular to each other.}\\ \text{Therefore, given points are the vertiaces of a right}\\ \text{angled triangle.}\end{array}$

Q.7 Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Ans

Angle of line with y-axis is 30°.
Angle of line with x-axis = 90° + 30° = 150°
Slope of line = tan150° = – √3

Q.8 Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.

Ans

$\begin{array}{l}\text{Let points are A}(\text{x},-\text{1}),\text{B}(\text{2},\text{1})\text{and C}(\text{4},\text{5}).\\ \mathrm{Slope}\text{of AB}=\frac{1+1}{2-\mathrm{x}}\\ =\frac{2}{2-\mathrm{x}}\\ \mathrm{Slope}\text{of BC}=\frac{5-1}{4-2}\\ =\frac{4}{2}\\ =2\\ \mathrm{Slope}\text{of CA}=\frac{-1-5}{4-\mathrm{x}}\\ \mathrm{If}\text{A, B and C are collinear.}\\ \text{Slope of AB}=\text{Slope of BC}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{2-\mathrm{x}}=2\\ \Rightarrow 2-\mathrm{x}=1\end{array}$ \begin{array}{l}\Rightarrow x=1\\ For\text{x}=\text{1, point A, B and C are collinear}\text{.}\end{array}

Q.9 Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Ans

$\begin{array}{l}\mathrm{The}\text{given points are letA}\left(-\text{2},-\text{1}\right),\text{B}\left(\text{4},0\right),\text{C}\left(\text{3},\text{3}\right)\text{and D}\left(-\text{3},\text{2}\right).\\ \mathrm{Slope}\text{of AB}=\frac{0+1}{4+2}\\ =\frac{1}{6}\\ \mathrm{Slope}\text{of BC}=\frac{3-0}{3-4}\\ =\frac{3}{-1}\\ =-3\\ \mathrm{Slope}\text{of CD}=\frac{2-3}{-3-3}\\ =\frac{-1}{-6}\\ =\frac{1}{6}\\ \mathrm{Slope}\text{of DA}=\frac{-1-2}{-2+3}\\ =\frac{-3}{1}\\ =-3\\ \mathrm{Since},\text{the slope of AB}=\text{the slope of CD}\\ \text{So, AB is parallel to CD.}\\ \mathrm{And},\text{the slope of BC}=\text{the slope of AD}\\ \text{So, BC is parallel to AD.}\\ \text{Therefore, ABCD is a parallelogram.}\end{array}$

Q.10 Find the angle between the x–axis and the line joining the points (3,–1) and (4,–2).

Ans

$\begin{array}{l}\text{m}=\text{Slope of}\left(\text{3,}–\text{1}\right)\text{and}\left(\text{4,}–\text{2}\right)\text{=}\frac{–\text{2}+\text{1}}{\text{4}–\text{3}}\\ \text{\hspace{0.17em} =}\frac{–\text{1}}{\text{1}}\\ \text{\hspace{0.17em} =}–\text{1}\\ \text{Let line makes θ angle with x-axis. Then,}\\ \text{tanθ}=–\text{1}\\ =\text{tan135°}\\ \Rightarrow \text{\hspace{0.17em}θ}=\text{135°}\\ \text{Thus, the angle between x-axis and line joining the given}\\ \text{point is 135°.}\end{array}$

Q.11

$\begin{array}{l}\mathbf{\text{The slope of a line is double of the slope of another line. If tangent of the angle between them}}\\ \mathbf{\text{is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{3}}}\mathbf{\text{, find the slopes of the lines.}}\end{array}$

Ans

Let the slope of first line, m₁ = 2m
Slope of other line, m₂ = m
Let angle between two lines be θ.

$\begin{array}{l}\mathrm{tan\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ \Rightarrow \frac{1}{3}=\left|\frac{\mathrm{m}-2\mathrm{m}}{1+2\mathrm{m}.\mathrm{m}}\right|\\ \Rightarrow \frac{1}{3}=\left|\frac{-\mathrm{m}}{1+2{\mathrm{m}}^2}\right|\\ \Rightarrow \frac{1}{3}=\frac{-\mathrm{m}}{1+2{\mathrm{m}}^2}\text{or \hspace{0.17em}}\frac{1}{3}=-\left(\frac{-\mathrm{m}}{1+2{\mathrm{m}}^2}\right)\\ \Rightarrow \frac{1}{3}=\frac{-\mathrm{m}}{1+2{\mathrm{m}}^2}\text{\hspace{0.17em}\hspace{0.17em}or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{3}=\frac{\mathrm{m}}{1+2{\mathrm{m}}^2}\\ \Rightarrow 2{\mathrm{m}}^2+1=-3\mathrm{m}\mathrm{or}2{\mathrm{m}}^2+1=3\mathrm{m}\\ \Rightarrow 2{\mathrm{m}}^2+3\mathrm{m}+1=0\mathrm{or}2{\mathrm{m}}^2-3\mathrm{m}+1=0\\ \Rightarrow 2{\mathrm{m}}^2+2\mathrm{m}+\mathrm{m}+1=0\mathrm{or}2{\mathrm{m}}^2-2\mathrm{m}-\mathrm{m}+1=0\\ \Rightarrow 2\mathrm{m}(\mathrm{m}+1)+1(\mathrm{m}+1)=0\mathrm{or}2\mathrm{m}(\mathrm{m}-1)-1(\mathrm{m}-1)=0\\ \Rightarrow (\mathrm{m}+1)(2\mathrm{m}+1)=0\mathrm{or}(\mathrm{m}-1)(2\mathrm{m}-1)=0\\ \Rightarrow \mathrm{m}=-1,-\frac{1}{2}\mathrm{orm}=-1,-\frac{1}{2}\\ \therefore {\mathrm{m}}_1=2(-1)\mathrm{or}{\text{m}}_{\text{1}}=2(-\frac{1}{2})\mathrm{or}{\mathrm{m}}_1=2\left(1\right)\mathrm{or}{\text{m}}_{\text{1}}=2\left(\frac{1}{2}\right)\\ {\mathrm{m}}_1=-2,-1,2,1{\text{and m}}_{\text{2}}=-1,-\frac{1}{2},-1,-\frac{1}{2}\\ \mathrm{Thus},\text{the slope of the lines are 1 and}2,\frac{1}{2}\text{\hspace{0.17em}and}1,-\text{1 and}-\text{2}\\ \text{or}-\frac{1}{2}\text{and}-1.\end{array}$

Q.12 A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).

Ans

$\begin{array}{l}\mathrm{Slope}\text{\hspace{0.17em}of line passing through}({\mathrm{x}}_1,{\mathrm{y}}_1)\text{and}(\mathrm{h},\mathrm{k})=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{h}-{\mathrm{x}}_1}\\ \mathrm{But},\text{\hspace{0.17em}the slope of line}=\mathrm{m}\\ \mathrm{then},\mathrm{m}=\frac{\mathrm{k}-{\mathrm{y}}_1}{\mathrm{h}-{\mathrm{x}}_1}\\ \Rightarrow \mathrm{k}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{h}-{\mathrm{x}}_1).\end{array}$

Q.13

$\mathbf{\text{If three points (h, 0), (a, b) and (0, k) lie on a line, show that}}\frac{\mathbf{\text{a}}}{\mathbf{\text{h}}}\mathbf{\text{+}}\frac{\mathbf{\text{b}}}{\mathbf{\text{k}}}\mathbf{\text{=1.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{the}\mathrm{three}\text{points}\mathrm{A}(\mathrm{h},0),\mathrm{B}(\mathrm{a},\mathrm{b})\mathrm{andC}(0,\mathrm{k})\text{lie on a line.}\\ \text{Then, slope of AB}=\text{slope of BC}\\ \Rightarrow \frac{\mathrm{b}-0}{\mathrm{a}-\mathrm{h}}=\frac{\mathrm{k}-\mathrm{b}}{0-\mathrm{a}}\\ \Rightarrow -\mathrm{ab}=(\mathrm{k}-\mathrm{b})(\mathrm{a}-\mathrm{h})\\ \Rightarrow -\mathrm{ab}=\mathrm{ak}-\mathrm{kh}-\mathrm{ab}+\mathrm{bh}\\ \Rightarrow 0=\mathrm{ak}-\mathrm{kh}+\mathrm{bh}\\ \Rightarrow \mathrm{ak}+\mathrm{bh}=\mathrm{kh}\\ \mathrm{Dividing}\text{both sides by kh, we get}\\ \frac{\mathrm{ak}}{\mathrm{kh}}+\frac{\mathrm{bh}}{\mathrm{kh}}=\frac{\mathrm{kh}}{\mathrm{kh}}\end{array}$ $\begin{array}{l}\Rightarrow \frac{\mathrm{a}}{\mathrm{h}}+\frac{\mathrm{b}}{\mathrm{k}}=1\\ \text{Hence proved.}\end{array}$

Q.14 Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?

Ans

\begin{array}{l}\text{Let the coordinates of point A and B are}\left(\text{1985},\text{92}\right)\text{and}\\ \left(\text{1995},\text{97}\right)\text{respectively}.\\ Then\text{, slope of line AB}=\frac{97-92}{1995-1985}\\ =\frac{5}{10}\\ =\frac{1}{2}\end{array} $\begin{array}{l}\mathrm{Let}\text{population in the year 2010}\\ =\mathrm{k}\\ \mathrm{Then},\text{coordinate of point C lying on line AB}\\ =(2010,\mathrm{k})\\ \mathrm{Slope}\text{of line BC}=\frac{\mathrm{k}-97}{2010-1995}\\ =\frac{\mathrm{k}-97}{25}\\ \mathrm{Since},\text{A, B and C lies on the same line. So,}\\ \text{Slope of line AB}=\mathrm{Slope}\text{of line BC}\\ \Rightarrow \frac{1}{2}=\frac{\mathrm{k}-97}{25}\\ \Rightarrow 25=2\mathrm{k}-194\\ \Rightarrow 25+194=2\mathrm{k}\\ \Rightarrow 219=2\mathrm{k}\\ \Rightarrow \mathrm{k}=\frac{219}{2}\\ =104.5\\ \mathrm{Thus},\text{slope of line is}\frac{1}{2}\text{and population in 2010 was 104.5 crores.}\end{array}$

Q.15 In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

1. Write the equations for the x-and y-axes.
2.

\text{Passing through the point (}–\text{4, 3) with slope}\frac{\text{1}}{\text{2}}\text{.}

3. Passing through (0, 0) with slope m.
4.

\text{Passing through (2, 2}\sqrt{\text{3}}\text{) and inclined with the x-axis at an angle of 75°}\text{.}

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
7. Passing through the points (–1, 1) and (2,– 4).
8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Ans

$\begin{array}{l}\mathrm{Slope}\left(\mathrm{m}\right)\text{of line}=\frac{1}{2}\\ \mathrm{Given}\text{point}=(-4,3)\\ \mathrm{Equation}\text{of line having slope m and passing through}\\ \text{the point}({\mathrm{x}}_1,{\mathrm{y}}_1)\\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)\\ \mathrm{y}-3=\frac{1}{2}(\mathrm{x}+4)\\ \Rightarrow 2(\mathrm{y}-3)=(\mathrm{x}+4)\\ \Rightarrow 2\mathrm{y}-6=\mathrm{x}+4\\ \Rightarrow \mathrm{x}-2\mathrm{y}+10=0\\ \mathrm{Thus},\text{x}-2\mathrm{y}+10=0\text{is the required equation of the line.}\end{array}$

$\begin{array}{l}\mathrm{Slope}\left(\mathrm{m}\right)\text{of line}=\mathrm{m}\\ \mathrm{Given}\text{point}=(0,0)\\ \mathrm{Equation}\text{of line having slope m and passing through}\\ \text{the point}({\mathrm{x}}_1,{\mathrm{y}}_1)\\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)\\ \mathrm{y}-0=\mathrm{m}(\mathrm{x}-0)\\ \Rightarrow \mathrm{y}=\mathrm{mx}\\ \mathrm{Thus},\mathrm{y}=\mathrm{mx}\text{is the required equation of the line.}\end{array}$

$\begin{array}{l}\mathrm{Slope}\left(\mathrm{m}\right)\text{of line}=\tan 75\mathrm{°}\\ =\tan (45\mathrm{°}+30\mathrm{°})\\ =\frac{\tan 45\mathrm{°}+\tan 30\mathrm{°}}{1-\tan 45\mathrm{°}\tan 30\mathrm{°}}\\ =\frac{1+\frac{1}{\sqrt{3}}}{1-1.\frac{1}{\sqrt{3}}}\\ =\frac{\sqrt{3}+1}{\sqrt{3}-1}\\ \mathrm{Given}\text{point}=(2,2\sqrt{3})\end{array}$ $\begin{array}{l}\mathrm{Equation}\text{of line having slope m and passing through}\\ \text{the point}({\mathrm{x}}_1,{\mathrm{y}}_1)\\ \Rightarrow \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)\\ \Rightarrow \mathrm{y}-2\sqrt{3}=\frac{\sqrt{3}+1}{\sqrt{3}-1}(\mathrm{x}-2)\\ \Rightarrow (\mathrm{y}-2\sqrt{3})(\sqrt{3}-1)=(\sqrt{3}+1)(\mathrm{x}-2)\\ \Rightarrow \mathrm{y}(\sqrt{3}-1)-2\sqrt{3}(\sqrt{3}-1)=(\sqrt{3}+1)\mathrm{x}-2(\sqrt{3}+1)\\ \Rightarrow (\sqrt{3}+1)\mathrm{x}-\mathrm{y}(\sqrt{3}-1)+2\sqrt{3}(\sqrt{3}-1)-2(\sqrt{3}+1)=0\\ \Rightarrow (\sqrt{3}+1)\mathrm{x}-(\sqrt{3}-1)\mathrm{y}+6-2\sqrt{3}-2\sqrt{3}-2=0\\ \Rightarrow (\sqrt{3}+1)\mathrm{x}-(\sqrt{3}-1)\mathrm{y}+4-4\sqrt{3}=0\\ \Rightarrow (\sqrt{3}+1)\mathrm{x}-(\sqrt{3}-1)\mathrm{y}+4(1-\sqrt{3})=0\\ \Rightarrow (\sqrt{3}+1)\mathrm{x}-(\sqrt{3}-1)\mathrm{y}=4(\sqrt{3}-1)\\ \mathrm{Thus},\mathrm{it}\text{is the required equation of the line.}\end{array}$

$\begin{array}{l}\mathrm{The}\text{coordinate of the point lying on x-axis at a distance of}\\ \text{3 units to the left of the origin}=(-3,0)\\ \mathrm{Slope}\text{of the line}=-2\\ \mathrm{Eqation}\text{of the line having slope}-\text{2 and passing through the}\\ \text{point}(-3,0)\text{is given as:}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}-0=-2(\mathrm{x}+3)[âˆ\mu \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)]\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=-2\mathrm{x}-6\\ \Rightarrow 2\mathrm{x}+\mathrm{y}+6=0\\ \mathrm{Which}\text{is the required equation of the given line.}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{The}\text{distance on y-axis of intersection point by line,}\left(\mathrm{c}\right)=2\\ \mathrm{Angle}\text{\hspace{0.17em}made by line with x-axis}=30\mathrm{°}\\ \mathrm{Slope}\text{of given line,m}=\tan 30\mathrm{°}\\ =\frac{1}{\sqrt{3}}\\ \mathrm{Equation}\text{\hspace{0.17em}of line having slope}\frac{1}{\sqrt{3}}\text{and having intercept on y-axis}\\ \text{is given as:}\\ \text{y}=\frac{1}{\sqrt{3}}\mathrm{x}+2[âˆ\mu \mathrm{y}=\mathrm{mx}+\mathrm{c}]\\ \Rightarrow \sqrt{3}\text{y}=\mathrm{x}+2\sqrt{3}\\ \Rightarrow \mathrm{x}-\sqrt{3}\mathrm{y}+2\sqrt{3}=0\\ \mathrm{Which}\text{is the required equation of the given line.}\end{array}$

$\mathrm{Equation}\text{of line passing through two point}({\mathrm{x}}_1,{\mathrm{y}}_1)\text{and}({\mathrm{x}}_2,{\mathrm{y}}_2)$ $\begin{array}{l}\mathrm{is}\text{\hspace{0.17em}given as: y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}(\mathrm{x}-{\mathrm{x}}_1)\\ \mathrm{Equation}\text{of line passing through two point}(-1,1)\mathrm{and}(2,-4)\\ \mathrm{is}:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}-1=\frac{-4-1}{2+1}(\mathrm{x}+1)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}y}-1=\frac{-5}{3}(\mathrm{x}+1)\\ \Rightarrow 3(\text{y}-1)=-5\mathrm{x}-5\\ \Rightarrow 5\mathrm{x}+3\mathrm{y}+2=0\\ \mathrm{Which}\text{is the required equation of the line.}\end{array}$

$\begin{array}{l}\text{Hence, the equation of the line having normal distance p from}\\ \text{the origin and angle}\mathrm{\omega }\text{which the normal makes with the positive}\\ \text{direction of x-axis is given by}\\ \text{x cos}\mathrm{\omega }\text{+ y sin}\mathrm{\omega }=\text{p}\\ \text{Here,}\mathrm{\omega }=30\mathrm{°}\text{and p}=5\text{units}\\ \text{So, equation of the line is:}\\ \text{\hspace{0.17em}\hspace{0.17em}x cos\hspace{0.17em}}30\mathrm{°}\text{+ y sin}30\mathrm{°}=\text{5}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} x}\left(\frac{\sqrt{3}}{2}\right)\text{+ y}\left(\frac{1}{2}\right)=\text{5}\\ \Rightarrow \sqrt{3}\mathrm{x}+\mathrm{y}=10\end{array}$ $\mathrm{Which}\text{is the required solution of the given equation.}$

Q.16 The vertices of Δ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

Ans

$\begin{array}{l}\mathrm{Coordinates}\text{of vertices of}\mathrm{\Delta PQR}\text{are:}\\ \text{P}(\text{2},\text{1}),\text{Q}(–\text{2},\text{3})\mathrm{}\text{and R}(\text{4},\text{5})\\ \mathrm{Median}\text{through point R bisects side PQ of}\mathrm{\Delta }\text{PQR.}\\ \text{Mid-point of PQ}=\{\frac{2+(-2)}{2},\frac{1+3}{2}\}\\ =(0,2)\\ \mathrm{Equation}\text{of median passing through}(4,5)\text{and}(0,2)\mathrm{is}:\\ \mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}(\mathrm{x}-{\mathrm{x}}_1)\\ \Rightarrow \mathrm{y}-5=\frac{2-5}{0-4}(\mathrm{x}-4)\\ \Rightarrow \mathrm{y}-5=\frac{-3}{-4}(\mathrm{x}-4)\\ \Rightarrow 4\mathrm{y}-20=3\mathrm{x}-12\\ 3\mathrm{x}-4\mathrm{y}+20-12=0\\ \Rightarrow 3\mathrm{x}-4\mathrm{y}+8=0\\ \mathrm{Therefore},\text{it is the equation of median passing}\\ \text{through vertex R.}\end{array}$

Q.17 Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Ans

$\begin{array}{l}\mathrm{Slope}\text{\hspace{0.17em}of the line through the points}(\text{2},\text{5})\text{and\hspace{0.17em}\hspace{0.17em}}(-\text{3},\text{\hspace{0.17em}\hspace{0.17em}6})=\frac{6-5}{-3-2}\\ =\frac{1}{-5}\\ \mathrm{Slope}\text{\hspace{0.17em}of the}\perp \text{on the line}=-\frac{1}{\left(\frac{1}{-5}\right)}\\ =5\\ \mathrm{Equation}\text{of the perpendicular on the line passing through the}\\ \text{point}(-3,5)\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}-5=5(\mathrm{x}+3)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}-5=5\mathrm{x}+15\\ \Rightarrow 5\mathrm{x}+15-\mathrm{y}+5=0\\ \Rightarrow 5\mathrm{x}-\mathrm{y}+20=0\\ \mathrm{Which}\text{is the required equation of line.}\end{array}$

Q.18 A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Ans

$\begin{array}{l}\mathrm{Slope}\text{of the line segment joining the points}(\text{1},0)\text{and}(\text{2},\text{3})\\ =\frac{3-0}{2-1}\\ =3\\ \mathrm{Slope}\text{of perpendicular to the line segment joining the points}\end{array}$ $\begin{array}{l}(1,0)\text{and}(2,3)=-\frac{1}{3}\\ \mathrm{Coordinate}\text{of the point dividing the line segment joining the}\\ \text{points}(\text{1},0)\text{and}(\text{2},\text{3})\text{in the ratio of 1:n is:}\\ (\frac{1.2+\mathrm{n}.1}{1+\mathrm{n}},\frac{1.3+\mathrm{n}.0}{1+\mathrm{n}})=(\frac{2+\mathrm{n}}{1+\mathrm{n}},\frac{3}{1+\mathrm{n}})\\ \mathrm{Equation}\text{of line intersecting line segment in the ratio of 1:n is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}-\frac{3}{1+\mathrm{n}}=-\frac{1}{3}(\mathrm{x}-\frac{2+\mathrm{n}}{1+\mathrm{n}})\\ \Rightarrow \frac{(1+\mathrm{n})\text{y}-3}{1+\mathrm{n}}=-\frac{1}{3}\left\{\frac{(1+\mathrm{n})\mathrm{x}-2-\mathrm{n}}{1+\mathrm{n}}\right\}\\ \Rightarrow 3(1+\mathrm{n})\text{y}-9=-(1+\mathrm{n})\mathrm{x}+2+\mathrm{n}\\ \Rightarrow (1+\mathrm{n})\mathrm{x}+3(1+\mathrm{n})\text{y}-2-\mathrm{n}-9=0\\ \Rightarrow (1+\mathrm{n})\mathrm{x}+3(1+\mathrm{n})\text{y}=\mathrm{n}+11\\ \mathrm{Which}\text{is the required equation of the line.}\end{array}$

Q.19 Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of line in the form of intercepts on axis is:}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\\ \mathrm{Since},\text{a}=\text{b. Then,}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{a}}=1...\left(\mathrm{i}\right)\end{array}$ $\begin{array}{l}\mathrm{Since},\text{this line passes through the point}(2,3).\mathrm{So},\\ \frac{2}{\mathrm{a}}+\frac{3}{\mathrm{a}}=1\\ \Rightarrow \frac{5}{\mathrm{a}}=1\\ \Rightarrow \mathrm{a}=5\\ \mathrm{Putting}\text{value of a in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{\mathrm{x}}{5}+\frac{\mathrm{y}}{5}=1\\ \Rightarrow \mathrm{x}+\mathrm{y}=5\\ \mathrm{Which}\text{is the required equation of line.}\end{array}$

Q.20 Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of line in the form of intercepts on axis is:}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\\ \mathrm{Since},\text{a}+\text{b}=\text{9. Then putting b}=\text{9}-\text{a, we get}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{9-\mathrm{a}}=1...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{this line passes through the point}(2,2).\mathrm{So},\\ \frac{2}{\mathrm{a}}+\frac{2}{9-\mathrm{a}}=1\\ \Rightarrow \frac{2(9-\mathrm{a})+2\mathrm{a}}{(9-\mathrm{a})\mathrm{a}}=1\end{array}$ $\begin{array}{l}\Rightarrow 18=9\mathrm{a}-{\mathrm{a}}^2\\ \Rightarrow {\mathrm{a}}^2-9\mathrm{a}+18=0\\ \Rightarrow (\mathrm{a}-6)(\mathrm{a}-3)=0\\ \Rightarrow \mathrm{a}=6,3\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}b}=\text{9}-\text{a}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} b}=3,6\\ \mathrm{Thus},\text{the equation of line is}\\ \frac{\mathrm{x}}{6}+\frac{\mathrm{y}}{3}=1\Rightarrow \mathrm{x}+2\mathrm{y}=6\\ \mathrm{or}\\ \frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{6}=1\Rightarrow 2\mathrm{x}+\mathrm{y}=6\end{array}$

Q.21

$\begin{array}{l}\mathbf{\text{Find equation of the line through the point (0, 2) making an}}\mathbf{angle}\mathbf{}\frac{\mathbf{\text{2π}}}{\mathbf{\text{3}}}\mathbf{\text{with the positive x-axis. Also,}}\\ \mathbf{\text{find the equation}}\mathbf{of}\mathbf{}\mathbf{line}\mathbf{}\mathbf{parallel}\mathbf{}\mathbf{to}\mathbf{}\mathbf{it}\mathbf{}\mathbf{and}\mathbf{}\mathbf{crossing}\mathbf{}\mathbf{the}\mathbf{}\mathbf{y}\mathbf{-}\mathbf{axis}\mathbf{}\mathbf{at}\mathbf{}\mathbf{a}\mathbf{}\mathbf{distance}\mathbf{\text{of 2 units below the origin.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Equation}\text{of line through the point}(0,2)\text{making an angle}\frac{2\mathrm{\pi }}{3}\\ \text{with the positive x-axis is}\\ \text{y}-2=\tan \left(\frac{2\mathrm{\pi }}{3}\right)(\mathrm{x}-0)\\ \Rightarrow \mathrm{y}-2=\tan (\mathrm{\pi }-\frac{\mathrm{\pi }}{3})\mathrm{x}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{y}-2=-\tan \left(\frac{\mathrm{\pi }}{3}\right)\mathrm{x}\\ \Rightarrow \mathrm{y}-2=-\sqrt{3}\mathrm{x}\\ \Rightarrow \sqrt{3}\mathrm{x}+\mathrm{y}-2=0\\ \mathrm{Slope}\text{of parallel line to}\sqrt{3}\mathrm{x}+\mathrm{y}-2=0,\text{is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m}=-\sqrt{3}\\ \mathrm{Equation}\text{of line of slope}-\sqrt{3}\text{and y-intercept}-\text{2 is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=-\sqrt{3}\mathrm{x}-2[\mathrm{y}=\mathrm{mx}+\mathrm{C}]\\ \sqrt{3}\mathrm{x}+\mathrm{y}+2=0\\ \mathrm{Therefore},\text{the required equations are:}\\ \sqrt{3}\mathrm{x}+\mathrm{y}-2=0\text{and}\sqrt{3}\mathrm{x}+\mathrm{y}+2=0.\end{array}$

Q.22 The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Ans

$\begin{array}{l}\text{Slope of perpendicular passing through}\left(\text{0,0}\right)\text{and}\left(\text{–2,9}\right)\text{=}\frac{9–0}{–2–0}\\ \text{=}–\frac{\text{9}}{\text{2}}\\ \text{Slope of perpendicular line}=–\frac{1}{\left(–\frac{9}{2}\right)}\\ \text{=}\frac{\text{2}}{\text{9}}\\ \text{Equationof line passing through}\left(–\text{2,9}\right)\text{and having slope}\frac{\text{2}}{\text{9}}\text{is}\\ \text{y}–\text{9}=\frac{\text{2}}{\text{9}}\left(\text{x}+\text{2}\right)\\ \text{9y}–\text{81}=\text{2x}+\text{4}\\ \Rightarrow \text{2x}–\text{9y}+\text{85}=\text{0}\\ \text{Which is required equation of line.}\end{array}$

Q.23 The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Ans

$\begin{array}{l}\text{Since, copper rod is a linear function of its Celsius temperature}\\ \text{C}.\text{In an experiment}\\ \text{L}=\text{124}.\text{942 when}\mathrm{}\text{C}=\text{2}0\text{and L}=\text{125}.\text{134 when C}=\text{11}0\\ \mathrm{Let}\mathrm{two}\text{points which satisfy the linear function between L and}\\ \text{C are}(124.942,20)\text{and}(125.134,110).\\ \mathrm{Let}\text{\hspace{0.17em}us consider that L is along x-axis and C is along y-axis.}\\ \text{So, the equation of line passing through the points}(124.942,20)\\ \text{and}(125.134,110)\text{in XY-plane.}\\ \text{L}-124.942=\frac{125.134-124.942}{110-20}(\mathrm{C}-20)\\ \Rightarrow \text{L}-124.942=\frac{0.192}{90}(\mathrm{C}-20)\\ \Rightarrow \text{\hspace{0.17em}L}=\frac{0.192}{90}(\mathrm{C}-20)+124.942\end{array}$ $\mathrm{Which}\text{is the required linear function of length a copper rod.}$

Q.24 The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?

Ans

$\begin{array}{l}\text{Let selling price per litre is along x-axis and demand along}\\ \text{y}–\text{axis. Two points}\left(\text{14,980}\right)\text{and}\left(\text{16,1220}\right)\text{are in XY}–\text{plane}\\ \text{and they satisfy the linear relation between selling price and}\\ \text{demand of milk. Then,}\\ \text{y}–\text{980}=\frac{\text{1220}–\text{980}}{\text{16}–\text{14}}\left(\text{x}–\text{14}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}–\text{980}=\frac{\text{240}}{\text{2}}\left(\text{x}–\text{14}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=\text{120}\left(\text{x}–\text{14}\right)\text{+980\hspace{0.17em} \hspace{0.17em}}...\left(\text{i}\right)\\ \text{When x}=\text{₹17/litre, from equation}\left(\text{i}\right)\text{:}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=\text{120}\left(\text{17}–\text{14}\right)+\text{980}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=\text{360}+\text{980}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1340\hspace{0.17em}litres}\\ \text{Thus, the owner of a milk store can sell 1340 litres of}\\ \text{milk each week at ₹17/litre.}\end{array}$

Q.25

$\begin{array}{l}\mathbf{\text{P(a, b) is the mid-point of a line segment between axes.}}\\ \mathbf{\text{Show that equation of the line is \hspace{0.17em}}}\frac{\mathbf{\text{x}}}{\mathbf{\text{a}}}\mathbf{\text{+}}\frac{\mathbf{\text{y}}}{\mathbf{\text{b}}}\mathbf{\text{=1.}}\end{array}$

Ans

$\begin{array}{l}\text{Let coordinates of A and B are}\left(\text{0,y}\right)\text{and}\left(\text{x,0}\right)\text{respectively.}\\ \text{Since, P}\left(\text{a,b}\right)\text{is mid point of AB.So,}\\ \text{a}=\frac{\text{0}+\text{x}}{\text{2}}\text{and b}=\frac{\text{y}+\text{0}}{\text{2}}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=\text{2a and \hspace{0.17em}y}=\text{2b}\\ \text{i.e., OB}=\text{2a and OA}=\text{2b}\\ \text{Equation of line in the intercept form is}\\ \frac{\text{x}}{\text{OB}}\text{+}\frac{\text{y}}{\text{OA}}\text{=1}\\ \Rightarrow \frac{\text{x}}{\text{2a}}\text{+}\frac{\text{y}}{\text{2b}}\text{=1}\\ \Rightarrow \frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{=2}\\ \text{which is the required equation.}\end{array}$

Q.26 Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Ans

$\begin{array}{l}\mathrm{Let}\text{coordinates of P and Q are}(0,\mathrm{y})\text{and}(\mathrm{x},0)\text{respectively.}\\ \text{Since, R}(\mathrm{h},\mathrm{k})\text{divides PQ in ratio 1:2, i.e., QR : QP}=\text{1:2. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} h}=\frac{2.\mathrm{x}+1.0}{1+2}\text{and k}=\frac{2.0+1.\mathrm{y}}{1+2}\\ \Rightarrow \mathrm{x}=\frac{3}{2}\mathrm{h}\text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=3\mathrm{k}\\ \mathrm{i}.\mathrm{e}.,\text{OQ}=\frac{3}{2}\mathrm{h}\text{and OP}=3\mathrm{k}\\ \mathrm{Equation}\text{of line in the intercept form is}\\ \frac{\mathrm{x}}{\mathrm{OQ}}+\frac{\mathrm{y}}{\mathrm{OP}}=1\\ \Rightarrow \frac{\mathrm{x}}{\frac{3}{2}\mathrm{h}}+\frac{\mathrm{y}}{3\mathrm{k}}=1\\ \Rightarrow \frac{2\mathrm{x}}{\mathrm{h}}+\frac{\mathrm{y}}{\mathrm{k}}=3\end{array}$ $\begin{array}{l}\Rightarrow 2\mathrm{kx}+\mathrm{hy}=3\mathrm{hk}\\ \text{Therefore, the required equation is}2\mathrm{kx}+\mathrm{hy}=3\mathrm{hk}.\end{array}$

Q.27 By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of line passing through the points\hspace{0.17em}\hspace{0.17em}}(3,0)\text{and}(-2,-2)\mathrm{is}:\\ \mathrm{y}-0=\frac{-2-0}{-2-3}(\mathrm{x}-3)\\ \Rightarrow \mathrm{y}=\frac{-2}{-5}(\mathrm{x}-3)\\ \Rightarrow 5\mathrm{y}=2\mathrm{x}-6\\ \Rightarrow 2\mathrm{x}-5\mathrm{y}-6=0\\ \mathrm{Let}\text{us check point}(8,2)\text{lies on the line or not.}\\ \mathrm{Putting}\text{x}=8\text{and y}=\text{2 in equation}\left(\mathrm{i}\right),\mathrm{we}\text{\hspace{0.17em}get}\\ \text{L.H.S.}=2\mathrm{x}-5\mathrm{y}-6\\ =2\left(8\right)-5\left(2\right)-6\\ =16-10-6\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \Rightarrow \text{Point}(8,2)\text{lies on the line\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}-5\mathrm{y}-6=0.\\ \mathrm{Therefore},\text{points}(3,0)\text{,}(-2,-2)\text{and\hspace{0.17em}\hspace{0.17em}}(8,2)\text{are collinear.}\end{array}$

Q.28 Reduce the following equations into slope – intercept form and find their slopes and the y – intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y – 5 = 0,
(iii) y = 0.

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{The}\text{given equation is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}+\text{7y}=0\\ \Rightarrow \text{y}=-\frac{1}{7}\mathrm{x}+0\\ \mathrm{Which}\text{is in the form of y}=\mathrm{mx}+\mathrm{c},\mathrm{where}\mathrm{m}=-\frac{1}{7}\\ \text{and C}=\text{0.}\\ \left(\text{ii}\right)\mathrm{The}\text{given equation is:}\\ \text{6x}+\text{3y}-\text{5}=0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 3y}=-6\mathrm{x}+5\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}=-2\mathrm{x}+\frac{5}{3}\\ \mathrm{Which}\text{is in the form of y}=\mathrm{mx}+\mathrm{c},\mathrm{where}\mathrm{m}=-2\\ \text{and C}=\frac{5}{3}\text{.}\\ \left(\mathrm{iii}\right)\mathrm{The}\text{given equation is:}\\ \mathrm{y}=0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=0.\mathrm{x}+0\\ \mathrm{Which}\text{is in the form of y}=\mathrm{mx}+\mathrm{c},\mathrm{where}\mathrm{m}=0\\ \text{and C}=0.\end{array}$

Q.29 Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0.

Ans

$\begin{array}{l}\left(\text{i}\right)\mathrm{The}\text{given equation is:}\\ \text{3}\mathrm{x}+\text{2}\mathrm{y}-\text{12}=0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\mathrm{x}+\text{2}\mathrm{y}=12\\ \mathrm{Dividing}\text{both sides by 12, we get}\\ \frac{\text{3}\mathrm{x}}{12}+\frac{\text{2}\mathrm{y}}{12}=1\\ \Rightarrow \frac{\mathrm{x}}{4}+\frac{\mathrm{y}}{6}=1\\ \mathrm{Which}\text{is in the intercept form and x,y- intercepts are 4 and}\\ \text{6 respectively.}\\ \left(\text{ii}\right)\mathrm{The}\text{given equation is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4}\mathrm{x}-\text{3}\mathrm{y}=\text{6}\mathrm{}\\ \mathrm{Dividing}\text{both sides by 6, we get}\\ \frac{\text{4}\mathrm{x}}{6}-\frac{\text{3}\mathrm{y}}{6}=\frac{\text{6}}{6}\mathrm{}\\ \Rightarrow \frac{\mathrm{x}}{\left(\frac{3}{2}\right)}-\frac{\mathrm{y}}{2}=1\mathrm{}\\ \Rightarrow \frac{\mathrm{x}}{\left(\frac{3}{2}\right)}+\frac{\mathrm{y}}{-2}=1\\ \mathrm{Which}\text{is in the intercept form and x,y- intercepts are}\frac{3}{2}\text{and}\\ -2\text{respectively.}\\ \left(\text{iii}\right)\mathrm{The}\text{given equation is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\mathrm{y}+\text{2}=0\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{y}=-\frac{2}{3}\\ \Rightarrow \frac{\mathrm{y}}{(-\frac{2}{3})}=1\\ \mathrm{Here},\text{y-intercept is}-\frac{2}{3}\text{and no intercept with x-axis.}\end{array}$

Q.30

$\begin{array}{l}\mathbf{\text{Reduce the following equations into normal form. Find theirperpendicular distances from the origin}}\\ \mathbf{\text{and angle betweenperpendicular and the positive x-axis.}}\\ \mathbf{\text{(i) x –}}\sqrt{\mathbf{\text{3}}}\mathbf{\text{y + 8 = 0,}}\\ \mathbf{\text{(ii) y – 2 = 0,}}\\ \mathbf{\text{(iii) x}}\mathbf{–}\mathbf{\text{y = 4.}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{x}-\sqrt{3}\mathrm{y}+8=0\\ -\mathrm{x}+\sqrt{3}\mathrm{y}=8\\ \mathrm{Dividing}\text{both sides by}\sqrt{1^2+{(-\sqrt{3})}^2}=2,\text{we get}\\ -\frac{1}{2}\mathrm{x}+\frac{\sqrt{3}}{2}\mathrm{y}=\frac{8}{2}\\ \Rightarrow \mathrm{xcos}\frac{2\mathrm{\pi }}{3}+\mathrm{ysin}\frac{2\mathrm{\pi }}{3}=4\\ \mathrm{Which}\text{is in the normal form i.e.,}\mathrm{xcos\omega }+\mathrm{ysin\omega }=\mathrm{p},\\ \mathrm{where}\text{perpendicular distance}\left(\mathrm{p}\right)=4\\ \mathrm{and}\text{angle between perpendicular}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and x-axis}\left(\mathrm{\omega }\right)=\frac{2\mathrm{\pi }}{3}\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{y}-2=0\\ \Rightarrow 0.\mathrm{x}+\mathrm{y}=2\\ \mathrm{Dividing}\text{both sides by}\sqrt{0^2+{\left(1\right)}^2}=1,\text{we get}\\ \Rightarrow \frac{0}{1}\mathrm{x}+\frac{\mathrm{y}}{1}=2\\ \Rightarrow \mathrm{x}\cos 90°+\mathrm{y}\sin 90°=2\\ \mathrm{Which}\text{is in the normal form i.e.,}\mathrm{x}\mathrm{cos\omega }+\mathrm{y}\mathrm{sin\omega }=\mathrm{p},\\ \mathrm{where}\text{perpendicular distance}\left(\mathrm{p}\right)=2\\ \mathrm{and}\text{angle between perpendicular}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and x-axis}\left(\mathrm{\omega }\right)=90°\\ \left(\mathrm{iii}\right)\mathrm{x}-\mathrm{y}=4\\ \mathrm{Dividing}\text{both sides by}\sqrt{1^2+{(-1)}^2}=\sqrt{2},\text{we get}\\ \Rightarrow \frac{1}{\sqrt{2}}\mathrm{x}-\frac{\mathrm{y}}{\sqrt{2}}=\frac{4}{\sqrt{2}}\\ \Rightarrow \mathrm{xcos}45°-\mathrm{y}\sin 45°=2\sqrt{2}\\ \Rightarrow \mathrm{xcos}(360°-45°)+\mathrm{y}\sin (360°-45°)=2\sqrt{2}\\ \Rightarrow \mathrm{xcos}(315°)+\mathrm{y}\sin (315°)=2\sqrt{2}\\ \mathrm{Which}\text{is in the normal form i.e.,}\mathrm{xcos}\mathrm{\omega }+\mathrm{y}\sin \mathrm{\omega }=\mathrm{p},\\ \mathrm{where}\text{perpendicular distance}\left(\mathrm{p}\right)=2\sqrt{2}\\ \mathrm{and}\text{angle between perpendicular}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and x-axis}\left(\mathrm{\omega }\right)=315°\end{array}$

Q.31 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Ans

Q.32

$\mathbf{\text{Find the points on the x-axis, whose distances from the line}}\frac{\mathbf{\text{x}}}{\mathbf{\text{3}}}\mathbf{\text{+}}\frac{\mathbf{\text{y}}}{\mathbf{\text{4}}}\mathbf{\text{=1\hspace{0.17em}\hspace{0.17em}are 4 units.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{the point on x-axis be}(\mathrm{a},0).\\ \text{Distance of the line}\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{4}=1\text{from the point}(\mathrm{a},0),\\ \mathrm{d}=\left|\frac{\frac{\mathrm{a}}{3}+\frac{0}{4}-1}{\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{4}\right)}^2}}\right|\end{array}$ $\begin{array}{l}\Rightarrow 4=\left|\frac{\frac{\mathrm{a}}{3}+\frac{0}{4}-1}{\sqrt{\left(\frac{1}{9}\right)+\left(\frac{1}{16}\right)}}\right|\\ \Rightarrow \pm 4=\frac{\frac{\mathrm{a}}{3}-1}{\sqrt{\left(\frac{16+9}{9\times 16}\right)}}\\ \Rightarrow \pm 4\left(\frac{5}{12}\right)=\frac{\mathrm{a}}{3}-1\\ \Rightarrow \pm \left(\frac{5}{3}\right)+1=\frac{\mathrm{a}}{3}\\ \mathrm{Taking}(+)\text{sign,}\\ \frac{5+3}{3}=\frac{\mathrm{a}}{3}\\ \Rightarrow \mathrm{a}=8\\ \mathrm{So},\text{the point on x-axis is}(8,0).\\ \mathrm{Taking}(-)\text{sign,}\\ \frac{-5+3}{3}=\frac{\mathrm{a}}{3}\\ \Rightarrow \mathrm{a}=-2\\ \mathrm{So},\text{the point on x-axis is}(-2,0).\\ \mathrm{Thus},\text{the distance of line from point}(-2,0)\text{and}(8,0)\text{is 4.}\end{array}$

Q.33 Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l (x + y) + p = 0 and l (x + y) – r = 0.

Ans

$\begin{array}{l}\left(\text{i}\right)\mathrm{Given}\text{parallel lines are:}\\ 15\mathrm{x}+8\mathrm{y}-34=0\mathrm{and}15\mathrm{x}+8\mathrm{y}+31=0\\ \text{Distance between two parallel lines,}\\ \text{d}=\left|\frac{{\mathrm{C}}_1-{\mathrm{C}}_2}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\right|\\ \mathrm{where},\text{A}=15,\mathrm{B}=8,{\mathrm{C}}_1=-34{\text{and C}}_{\text{2}}=31\\ \therefore \text{\hspace{0.17em}\hspace{0.17em} d}=\left|\frac{-34-31}{\sqrt{{15}^2+8^2}}\right|\\ =\left|\frac{-65}{\sqrt{289}}\right|\\ =\frac{65}{17}\text{units}\\ \text{Thus, the distance between two given parallel}\\ \text{lines is}\frac{65}{17}\text{units.}\\ \left(\text{ii}\right)\mathrm{Given}\text{parallel lines are:}\\ \mathrm{l}\text{(x}+\text{y)}+\text{p = 0 and}\mathrm{l}\text{(x}+\text{y)}-\text{r = 0}\\ \text{Distance between two parallel lines,}\\ \text{d}=\left|\frac{{\mathrm{C}}_1-{\mathrm{C}}_2}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\right|\\ \mathrm{where},\text{A}=\mathrm{l},\mathrm{B}=\mathrm{l},{\mathrm{C}}_1=\mathrm{p}{\text{and C}}_{\text{2}}=-\text{\hspace{0.17em}r}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em} d}=\left|\frac{\mathrm{p}+\mathrm{r}}{\sqrt{{\mathrm{l}}^2+{\mathrm{l}}^2}}\right|\end{array}$ $\begin{array}{l}=\left|\frac{\mathrm{p}+\mathrm{r}}{\sqrt{2{\mathrm{l}}^2}}\right|\\ =\frac{1}{\sqrt{2}}\left|\frac{\mathrm{p}+\mathrm{r}}{\mathrm{l}}\right|\text{units}\\ \text{Thus, the distance between two given parallel}\\ \text{lines is}\frac{1}{\sqrt{2}}\left|\frac{\mathrm{p}+\mathrm{r}}{\mathrm{l}}\right|\text{units.}\end{array}$

Q.34 Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\text{equation of the line parallel to the line}\mathrm{}\text{3}\mathrm{x}-\text{4}\mathrm{y}+\text{2}=0\text{is}\\ \text{3}\mathrm{x}-\text{4}\mathrm{y}+\mathrm{\lambda }=0\text{and it is passing through the point}(-2,3),\mathrm{then}\\ \text{3}(-2)-\text{4}\left(3\right)+\mathrm{\lambda }=0\\ \Rightarrow -6-12+\mathrm{\lambda }=0\\ \Rightarrow -18+\mathrm{\lambda }=0\\ \Rightarrow \mathrm{\lambda }=18\\ \mathrm{Thus},\text{parallel line is 3}\mathrm{x}-\text{4}\mathrm{y}+18=0.\end{array}$

Q.35 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Ans

$\begin{array}{l}\mathrm{The}\text{given equation of line\hspace{0.17em} is x}-\text{7y}+\text{5}=0.\\ \mathrm{Then},\text{given equation of line can be written as:}\\ \text{y}=\frac{1}{7}\mathrm{x}+\frac{5}{7},\text{comparing with}\mathrm{y}=\mathrm{mx}+\mathrm{c},\text{we have}\\ \text{m}=\frac{1}{7}\end{array}$ $\begin{array}{l}\mathrm{Slope}\text{of perpendicular line,}\\ \text{M}=-\frac{1}{\mathrm{m}}=-7\\ \mathrm{Equation}\text{of perpendicular line is:}\\ \text{\hspace{0.17em}y}=\text{Mx}+\text{C’}\\ \text{\hspace{0.17em}y}=-7\mathrm{x}+\mathrm{C}‘\\ \text{Since, perpendicular has x-intercept 3. So,}\\ \text{perpendicular passes through the point}(3,0).\\ \text{\hspace{0.17em} 0}=-7\left(3\right)+\mathrm{C}‘\\ \Rightarrow \mathrm{C}‘=21\\ \mathrm{Thus},\text{the equation of perpendicular line is:}\\ \text{\hspace{0.17em} y}=-7\mathrm{x}+21\text{or 7x}+\text{y}-\text{21}=\text{0}\end{array}$

Q.36

$\mathbf{\text{Find angles between the lines}}\sqrt{\mathbf{\text{3}}}\mathbf{\text{\hspace{0.17em}x + y = 1and x +}}\sqrt{\mathbf{\text{3}}}\mathbf{\text{y = 1.}}$

Ans

$\begin{array}{l}\mathrm{Thegiven}\mathrm{linesare}\\ \sqrt{3}\mathrm{x}+\mathrm{y}=1\Rightarrow \mathrm{y}=-\sqrt{3}\mathrm{x}+1...\left(\mathrm{i}\right)\\ \mathrm{and}\mathrm{x}+\sqrt{3}\mathrm{y}=1\Rightarrow \mathrm{y}=-\frac{1}{\sqrt{3}}\mathrm{x}+\frac{1}{\sqrt{3}}...\left(\mathrm{ii}\right)\\ \mathrm{The}\text{slope of equation}\left(\mathrm{i}\right),{\mathrm{m}}_1=-\sqrt{3}\\ \mathrm{The}\text{slope of equation}\left(\mathrm{ii}\right),{\mathrm{m}}_2=-\frac{1}{\sqrt{3}}\\ \mathrm{Angle}\text{between two lines is}\mathrm{\theta }.\\ \mathrm{Then},\\ \mathrm{tan\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\end{array}$ $\begin{array}{l}=\left|\frac{-\frac{1}{\sqrt{3}}-(-\sqrt{3})}{1+(-\sqrt{3})(-\frac{1}{\sqrt{3}})}\right|\\ =\left|\frac{\frac{-1+3}{\sqrt{3}}}{1+1}\right|\\ =\left|\frac{\frac{2}{\sqrt{3}}}{2}\right|\\ \mathrm{tan\theta }=\pm \frac{1}{\sqrt{3}}\\ \mathrm{tan\theta }=\frac{1}{\sqrt{3}}\text{or}-\frac{1}{\sqrt{3}}\\ \Rightarrow \mathrm{tan\theta }=\tan 30\mathrm{°}\text{or tan150°}\\ \Rightarrow \mathrm{\theta }=30\mathrm{°}\text{or 150°}\\ \text{Thus, the angle between two lines is 30° or 150°}.\end{array}$

Q.37 The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y +19 = 0. at right angle. Find the value of h.

Ans

$\begin{array}{l}\mathrm{Slope}\text{of the line through the points}(\text{h},\text{3})\text{and}(\text{4},\text{1}),{\mathrm{m}}_1=\frac{1-3}{4-\mathrm{h}}\\ \mathrm{Slope}\text{of the line 7x}-\text{9y}+\text{19}=0,{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m}}_{\text{2}}=-\frac{7}{-9}=\frac{7}{9}\end{array}$ $\begin{array}{l}\mathrm{Since},\text{both lines are perpendicular to each other, so}\\ {\text{m}}_{\text{1}}\times {\text{m}}_{\text{2}}=-1\\ \frac{1-3}{4-\mathrm{h}}\times \frac{7}{9}=-1\\ \Rightarrow -14=-(36-9\mathrm{h})\\ \Rightarrow 14=36-9\mathrm{h}\\ \Rightarrow \mathrm{h}=\frac{36-14}{9}\\ =\frac{22}{9}\\ \mathrm{Thus},\text{the value of h is}\frac{22}{9}.\end{array}$

Q.38 Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x –x₁) + B (y – y₁) = 0.

Ans

$\begin{array}{l}\mathrm{Slope}\text{of the line Ax}+\text{By}+\text{C}=0\text{, m}=-\frac{\mathrm{A}}{\mathrm{B}}\\ \mathrm{Equation}\text{of parallel line to given line is:}\\ \text{y}=-\frac{\mathrm{A}}{\mathrm{B}}\mathrm{x}+\mathrm{C}‘...\left(\mathrm{i}\right)\\ \mathrm{Equation}\left(\mathrm{i}\right)\text{is passing through the point}({\mathrm{x}}_1,{\mathrm{y}}_1),\\ \text{then from equation}\left(\mathrm{i}\right)\\ {\text{y}}_{\text{1}}=-\frac{\mathrm{A}}{\mathrm{B}}{\mathrm{x}}_1+\mathrm{C}‘\\ \Rightarrow \mathrm{C}‘=\frac{\mathrm{A}}{\mathrm{B}}{\mathrm{x}}_1+{\text{y}}_{\text{1}}\end{array}$ $\begin{array}{l}\mathrm{Putting}\text{value of C’ in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{y}=-\frac{\mathrm{A}}{\mathrm{B}}\mathrm{x}+\frac{\mathrm{A}}{\mathrm{B}}{\mathrm{x}}_1+{\text{y}}_{\text{1}}\\ \Rightarrow \mathrm{B}\text{y}=-\mathrm{Ax}+{\mathrm{Ax}}_1+\mathrm{B}{\text{y}}_{\text{1}}\\ \mathrm{A}(\mathrm{x}-{\mathrm{x}}_1)+\mathrm{B}(\mathrm{y}-{\mathrm{y}}_1)=0\\ \mathrm{Which}\text{is the required equation of parallel line.}\end{array}$

Q.39 Two lines passing through the point (2, 3) intersects each other at an angle of 60^°. If slope of one line is 2, find equation of the other line.

Ans

$\begin{array}{l}\mathrm{Let}{\text{slope of one line, m}}_{\text{1}}=2\text{and}\\ {\text{slope of other line, m}}_{\text{2}}=\text{m.}\\ \text{Angle between two lines}=60°\\ \mathrm{Since},\text{tan}\mathrm{\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60°=\left|\frac{\mathrm{m}-2}{1+2\mathrm{m}}\right|\\ \Rightarrow \sqrt{3}=\left|\frac{\mathrm{m}-2}{1+2\mathrm{m}}\right|\\ \Rightarrow \pm \sqrt{3}=\frac{\mathrm{m}-2}{1+2\mathrm{m}}\end{array}$ $\begin{array}{l}\mathrm{For}(+)\mathrm{ve}\text{sign:}\\ \sqrt{3}=\frac{\mathrm{m}-2}{1+2\mathrm{m}}\\ \sqrt{3}(1+2\mathrm{m})=\mathrm{m}-2\\ \sqrt{3}+2\sqrt{3}\mathrm{m}=\mathrm{m}-2\\ (2\sqrt{3}-1)\mathrm{m}=-(2+\sqrt{3})\\ \mathrm{m}=-\frac{(2+\sqrt{3})}{(2\sqrt{3}-1)}\\ \mathrm{Equation}\text{of line passing through the point}(2,3):\\ \mathrm{y}-3=-\frac{(2+\sqrt{3})}{(2\sqrt{3}-1)}(\mathrm{x}-2)\\ (\mathrm{y}-3)(2\sqrt{3}-1)=-(2+\sqrt{3})\mathrm{x}+2(2+\sqrt{3})\\ (2+\sqrt{3})\mathrm{x}+(2\sqrt{3}-1)\mathrm{y}=2(2+\sqrt{3})+3(2\sqrt{3}-1)\\ (2+\sqrt{3})\mathrm{x}+(2\sqrt{3}-1)\mathrm{y}=4+2\sqrt{3}+6\sqrt{3}-3\\ (\sqrt{3}+2)\mathrm{x}+(2\sqrt{3}-1)\mathrm{y}=1+8\sqrt{3}\\ \mathrm{For}(-)\mathrm{ve}\text{sign:}\\ -\sqrt{3}=\frac{\mathrm{m}-2}{1+2\mathrm{m}}\\ -\sqrt{3}(1+2\mathrm{m})=\mathrm{m}-2\\ -\sqrt{3}-2\sqrt{3}\mathrm{m}=\mathrm{m}-2\\ 2-\sqrt{3}=(2\sqrt{3}+1)\mathrm{m}\end{array}$ $\begin{array}{l}\mathrm{m}=\frac{2-\sqrt{3}}{(2\sqrt{3}+1)}\\ \mathrm{Equation}\text{of line passing through the point}(2,3):\\ \mathrm{y}-3=\frac{2-\sqrt{3}}{(2\sqrt{3}+1)}(\mathrm{x}-2)\\ \Rightarrow (2\sqrt{3}+1)\mathrm{y}-3(2\sqrt{3}+1)=(2-\sqrt{3})\mathrm{x}-4+2\sqrt{3}\\ \Rightarrow (\sqrt{3}-2)\mathrm{x}+(2\sqrt{3}+1)\mathrm{y}=6\sqrt{3}+3-4+2\sqrt{3}\\ \Rightarrow (\sqrt{3}-2)\mathrm{x}+(2\sqrt{3}+1)\mathrm{y}=8\sqrt{3}-1\\ \mathrm{Therefore},\text{the equation of other line is}\\ (\sqrt{3}+2)\mathrm{x}+(2\sqrt{3}-1)\mathrm{y}=1+8\sqrt{3}\text{\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}(\sqrt{3}-2)\mathrm{x}+(2\sqrt{3}+1)\mathrm{y}=8\sqrt{3}-1.\end{array}$

Q.40 Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Ans

$\begin{array}{l}\mathrm{Mid}-\mathrm{point}\text{of}(3,4)\text{and}(-1,2)=(\frac{3-1}{2},\frac{4+2}{2})\\ =(1,3)\\ \mathrm{Slope}\text{of line joining the points}(3,4)\text{and}(-1,2)\\ =\frac{2-4}{-1-3}\\ =\frac{-2}{-4}\\ =\frac{1}{2}\end{array}$ $\begin{array}{l}\mathrm{Slope}\text{of right bisector of line}=-\frac{1}{\left(\frac{1}{2}\right)}\\ =-2\\ \mathrm{Equation}\text{of line passing throgh the point}(1,3)\text{and}\\ \text{having slope}-\text{2.}\\ \text{y}-\text{3}=-2(\mathrm{x}-1)\\ \text{y}-\text{3}=-2\mathrm{x}+2\\ 2\mathrm{x}+\mathrm{y}=5\\ \mathrm{Thus},\text{the equation of right bisector is}2\mathrm{x}+\mathrm{y}=5.\end{array}$

Q.41 Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Ans

$\begin{array}{l}\mathrm{Given}\text{line is:}\\ \text{\hspace{0.17em} 3}\mathrm{x}-\text{4}\mathrm{y}-\text{16}=0\\ \Rightarrow \text{3}\mathrm{x}-\text{4}\mathrm{y}=\text{16}...\left(\mathrm{i}\right)\\ \mathrm{Slope}\text{of the line 3}\mathrm{x}-\text{4}\mathrm{y}-\text{16}=0,\mathrm{m}=-\frac{3}{-4}[âˆ\mu \mathrm{m}=-\frac{\mathrm{A}}{\mathrm{B}}]\\ \mathrm{m}=\frac{3}{4}\\ \mathrm{Slope}\text{of perpendicular to the line}=-\frac{1}{\left(\frac{3}{4}\right)}\\ =-\frac{4}{3}\end{array}$ $\begin{array}{l}\mathrm{Equation}\text{of perpendicular through the point}(-1,3)\mathrm{is}\\ \mathrm{y}-3=-\frac{4}{3}(\mathrm{x}+1)\\ \Rightarrow 3\mathrm{y}-9=-4\mathrm{x}-4\\ \Rightarrow 4\mathrm{x}+3\mathrm{y}=9-4\\ \Rightarrow 4\mathrm{x}+3\mathrm{y}=5...\left(\mathrm{ii}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{x}=\frac{68}{25}\text{and y}=-\frac{49}{25}\\ \mathrm{Thus},\text{the coordinates of the foot of perpendicular}\\ \text{is}(\frac{68}{25},-\frac{49}{25}).\end{array}$

Q.42 The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

Ans

$\begin{array}{l}\mathrm{Slope}\text{of line from origin}(0,0)\text{to}(-1,2),{\mathrm{m}}_1=\frac{2-0}{-1-0}\\ =-2\\ \mathrm{Slope}\text{of the perpendicular line}=\mathrm{m}\\ \mathrm{Then},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m}}_{\text{1}}{\text{m}}_{\text{2}}=-\text{1}\\ \Rightarrow -2\times \mathrm{m}=-1\\ \Rightarrow \mathrm{m}=\frac{1}{2}\\ \mathrm{Putting}\text{m}=\frac{1}{2}\text{in equation, y = mx}+\text{c, we get}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}y}=\frac{1}{2}\mathrm{x}+\mathrm{c}...\left(\mathrm{i}\right)\\ \mathrm{The}\text{point}(-\text{1},\text{2})\text{lies on line}\left(\mathrm{i}\right),\mathrm{so}\\ \text{\hspace{0.17em}2}=\frac{1}{2}(-1)+\mathrm{c}\\ \mathrm{c}=2+\frac{1}{2}\\ =\frac{5}{2}\\ \mathrm{Thus},\text{the values of m and c are}\frac{1}{2}\text{and}\frac{5}{2}\text{respectively.}\end{array}$

Q.43 If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p² + 4q² = k².

Ans

$\begin{array}{l}\mathrm{The}\text{given lines are:}\mathrm{}\text{\hspace{0.17em}\hspace{0.17em}xcos}\mathrm{\theta }-\text{ysin}\mathrm{\theta }=\text{kcos 2}\mathrm{\theta }...\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}xsec}\mathrm{\theta }+\text{ycosec}\mathrm{\theta }=\text{k\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{p}=\mathrm{Length}\text{of perpendicular from origin to line}\left(\mathrm{i}\right)\\ =\left|\frac{\left(0\right)\text{cos}\mathrm{\theta }-\left(0\right)\text{sin}\mathrm{\theta }-\text{kcos 2}\mathrm{\theta }}{\sqrt{{\cos }^2\mathrm{\theta }+{\text{sin}}^2\mathrm{\theta }}}\right|\\ \mathrm{p}=\text{kcos 2}\mathrm{\theta }\\ \mathrm{q}=\mathrm{Length}\text{of perpendicular from origin to line}\left(\mathrm{ii}\right)\\ =\left|\frac{\left(0\right)\text{sec}\mathrm{\theta }+\left(0\right)\text{cosec}\mathrm{\theta }-\text{k}}{\sqrt{{\text{sec}}^2\mathrm{\theta }+{\text{cosec}}^2\mathrm{\theta }}}\right|\end{array}$ $\begin{array}{l}=\frac{\mathrm{k}}{\sqrt{\frac{{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }\times {\sin }^2\mathrm{\theta }}}}\\ \mathrm{q}=\frac{\mathrm{ksin\theta }\mathrm{cos\theta }}{\sqrt{{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}}\\ =\mathrm{ksin\theta cos\theta }\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\text{p}}^{\text{2}}+{\text{4q}}^{\text{2}}\\ ={\left(\text{kcos 2}\mathrm{\theta }\right)}^2+4{(\mathrm{ksin\theta }\mathrm{cos\theta })}^2\\ ={\mathrm{k}}^2{\left(\text{cos 2}\mathrm{\theta }\right)}^2+{\mathrm{k}}^2{(2\mathrm{sin\theta }\mathrm{cos\theta })}^2\\ ={\mathrm{k}}^2({\text{cos}}^2\text{2}\mathrm{\theta }+{\sin }^{2}2\mathrm{\theta })\\ ={\mathrm{k}}^2\times 1\\ ={\mathrm{k}}^2=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.44 In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Ans

$\begin{array}{l}\mathrm{Slope}\text{of BC}=\frac{2+1}{1-4}\\ =-\frac{3}{3}\\ =-1\\ \mathrm{Slope}\text{of perpendicular on BC}=-\frac{1}{-1}\\ =1\\ \mathrm{Equation}\text{of altitude on BC through A}(2,3)\\ \mathrm{y}-3=1(\mathrm{x}-2)\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{y}-3=\mathrm{x}-2\\ \Rightarrow \mathrm{x}-\mathrm{y}=-3+2\\ \Rightarrow \mathrm{y}-\mathrm{x}=1\\ \mathrm{Equation}\text{of line through BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}+1=\frac{2+1}{1-4}(\mathrm{x}-4)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}+1=\frac{3}{-3}(\mathrm{x}-4)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}+1=-1(\mathrm{x}-4)\\ \Rightarrow \mathrm{x}+\mathrm{y}=3\\ \mathrm{Length}\text{of perpendicular from A}(2,3)\text{to BC}\\ \mathrm{d}=\left|\frac{2+3-3}{\sqrt{1^2+{(-1)}^2}}\right|\\ =\frac{2}{\sqrt{2}}\\ \mathrm{d}=\sqrt{2}\\ \mathrm{Thus},\text{equation of perpendicular is}\mathrm{y}-\mathrm{x}=1\text{and length of}\\ \text{perpendicular is}\sqrt{2}.\end{array}$

Q.45

$\begin{array}{l}\text{If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a}\\ \text{and b,then show that \hspace{0.17em}}\frac{\text{1}}{{\text{p}}^{\text{2}}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{2}}}\text{+}\frac{\text{1}}{{\text{b}}^{\text{2}}}\text{.}\end{array}$

Ans

$\mathrm{Equation}\text{of line having intercepts a and b with axis,}$ $\begin{array}{l}\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1...\left(\mathrm{i}\right)\\ \mathrm{p}=\mathrm{Length}\text{of perpendicular from origin to line}\left(\mathrm{i}\right)\\ =\left|\frac{\frac{0}{\mathrm{a}}+\frac{0}{\mathrm{b}}-1}{\sqrt{{\left(\frac{1}{\mathrm{a}}\right)}^2+{\left(\frac{1}{\mathrm{b}}\right)}^2}}\right|\\ \mathrm{p}=\left|\frac{1}{\sqrt{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}}}\right|\\ \mathrm{Squarring}\text{both sides, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} p}}^{\text{2}}={\left(\frac{1}{\sqrt{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}}}\right)}^2\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} p}}^{\text{2}}=\frac{1}{(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2})}\\ \Rightarrow \frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=\frac{1}{{\mathrm{p}}^2}\\ \mathrm{Therefore},\frac{1}{{\mathrm{p}}^2}=\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\text{\hspace{0.17em}\hspace{0.17em}has proved.}\end{array}$

Q.46 Find the equation of the line through the intersection of lines 3x + 4y = 7 and x – y + 2 = 0 and whose slope is 5.

Ans

The equation of any line through the point of intersection of the given lines is of the form

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 3x}+\text{4y}-\text{7}+\text{k}(\text{x}-\text{y}+\text{2})=0\\ \Rightarrow \mathrm{}(\text{3}+\text{k})\text{x}+(\text{4}–\text{k})\text{y}–\text{7}+\text{2k}=0...\left(\mathrm{i}\right)\\ \mathrm{Slope}\text{of line}\left(\mathrm{i}\right)=-\frac{3+\mathrm{k}}{4-\mathrm{k}}\\ \mathrm{Given}\text{slope of line}\left(\mathrm{i}\right)=5\\ \mathrm{Then},-\frac{3+\mathrm{k}}{4-\mathrm{k}}=5\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow -3-\mathrm{k}=20-5\mathrm{k}\\ \Rightarrow 4\mathrm{k}=23\\ \Rightarrow \mathrm{k}=\frac{23}{4}\\ \mathrm{Putting}\mathrm{k}=\frac{23}{4}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ (\text{3}+\frac{23}{4})\text{x}+(\text{4}-\frac{23}{4})\text{y}-\text{7}+\text{2}\times \frac{23}{4}=0\\ \Rightarrow 35\mathrm{x}-7\mathrm{y}+18=0\\ \mathrm{Therefore},\text{the required equation of line is}\\ 35\mathrm{x}-7\mathrm{y}+18=0.\end{array}$

Q.47 Find the equation of the line through the intersection of lines x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 0

Ans

$\begin{array}{l}\text{The equation of any line through the point of intersection of the given lines is of the form}\\ \text{x}+\text{2y}-\text{3}+\text{k}\left(\text{4x}-\text{y}+\text{7}\right)=0\\ \Rightarrow \left(\text{1}+4\text{k}\right)\text{x}+\left(\text{2}-\text{k}\right)\text{y}-\text{3}+\text{7k}=0...\left(\text{i}\right)\\ \mathrm{Slope}\text{of line}\left(\text{i}\right)=-\frac{\text{1}+4\text{k}}{\text{2}-\text{k}}\\ \mathrm{Equation}\text{of given line parallel to line}\left(\text{i}\right)\text{is:}\\ 5\text{x}+4\text{y}-20=0...\left(\mathrm{ii}\right)\\ \mathrm{Slope}\text{of line}\left(\mathrm{ii}\right)=-\frac{5}{4}\\ \mathrm{According}\text{to given condition:}\\ -\frac{\text{1}+4\text{k}}{\text{2}-\text{k}}=-\frac{5}{4}\\ \Rightarrow \frac{\text{1}+4\text{k}}{\text{2}-\text{k}}=\frac{5}{4}\\ \Rightarrow 4+16\text{k}=10-5\text{k}\\ \Rightarrow 21\text{k}=6\\ \Rightarrow \text{k}=\frac{6}{21}=\frac{2}{7}\\ \mathrm{Putting}\text{value of k in equation}\left(\text{i}\right),\text{we get}\\ \left(\text{1}+4\times \frac{2}{7}\right)\text{x}+\left(\text{2}-\frac{2}{7}\right)\text{y}-\text{3}+\text{7}\times \frac{2}{7}=0\\ \Rightarrow \frac{15}{7}\text{x}+\frac{12}{7}\text{y}-1=0\\ \Rightarrow 15\text{x}+12\text{y}-1=0\\ \mathrm{Therefore},\text{the required equation of line is}\\ 15\text{x}+12\text{y}-1=0.\end{array}$

Q.48 Find the equation of the line through the intersection of the lines 2x + 3y – 4 = 0 and x – 5y = 7 that has its x-intercept equal to – 4.

Ans

$\begin{array}{l}\text{The equation of any line through the point ofintersection of the given lines is of the form}\\ 2\text{x}+\text{3y}-\text{4}+\text{k}\left(\text{x}-5\text{y}-\text{7}\right)=0\\ \Rightarrow \left(\text{2}+\text{k}\right)\text{x}+\left(\text{3}-5\text{k}\right)\text{y}-\text{4}-\text{7k}=0...\left(\text{i}\right)\\ \mathrm{Intercept}\text{form of equation}\left(\text{i}\right)\text{is:}\\ \frac{\text{x}}{\left(\frac{\text{4}+\text{7k}}{\text{2}+\text{k}}\right)}+\frac{\text{y}}{\left(\frac{\text{4}+\text{7k}}{\text{3}-5\text{k}}\right)}=\text{1}\\ \text{Since, x-intercept of line}\left(\text{i}\right)=-4\\ \Rightarrow \frac{\text{4}+\text{7k}}{\text{2}+\text{k}}=-4\\ \Rightarrow \text{4}+\text{7k}=-8-4\text{k}\\ \Rightarrow 11\text{k}=-12\\ \Rightarrow \text{k}=-\frac{12}{11}\\ \mathrm{Putting}\text{value of k in equation}\left(\text{i}\right),\text{we get}\\ \left(\text{2}-\frac{12}{11}\right)\text{x}+\left(\text{3}-5\times -\frac{12}{11}\right)\text{y}-\text{4}-\text{7}\times -\frac{12}{11}=0\\ \Rightarrow \frac{10}{11}\text{x}+\frac{69}{11}\text{y}+\frac{40}{11}=0\\ \Rightarrow 10\text{x}+69\text{y}+40=0\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{line}\\ \text{is}\frac{10}{11}\text{x}+\frac{69}{11}\text{y}+\frac{40}{11}=0.\end{array}$

Q.49 Find the equation of the line through the intersection of 5x – 3y = 1 and 2x + 3y – 23 = 0 and perpendicular to the line 5x – 3y – 1 = 0.

Ans

$\begin{array}{l}\text{The equation of any line through the point of}\mathrm{}\\ \text{intersection of the given lines is of the form}\\ 5\text{x}-\text{3y}-\text{1}+\text{k}(\text{2x}+3\text{y}-\text{23})=0\\ \Rightarrow \mathrm{}(\text{5}+2\text{\hspace{0.17em}k})\text{x}+(-\text{3}+3\text{k})\text{y}-\text{1}-\text{23k}=0...\left(\mathrm{i}\right)\\ \mathrm{Slope}\text{of line}\left(\mathrm{i}\right)=-\frac{\text{5}+2\text{k}}{-\text{3}+3\text{k}}\\ \mathrm{Equation}\text{of given line perpendicular to line}\left(\mathrm{i}\right)\text{is:}\\ 5\mathrm{x}-3\mathrm{y}-1=0...\left(\mathrm{ii}\right)\\ \mathrm{Slope}\text{of line}\left(\mathrm{ii}\right)=-\frac{5}{-3}\\ =\frac{5}{3}\\ \mathrm{Since},\text{line}\left(\mathrm{i}\right)\text{and line}\left(\mathrm{ii}\right)\text{are perpendicular.So,}\\ -\frac{\text{5}+2\text{k}}{-\text{3}+3\text{k}}\times \frac{5}{3}=-1\\ \Rightarrow \frac{\text{5}+2\text{k}}{-\text{3}+3\text{k}}\times \frac{5}{3}=1\end{array}$ $\begin{array}{l}\Rightarrow 25+10\mathrm{k}=-9+9\mathrm{k}\\ \Rightarrow \mathrm{k}=-34\\ \mathrm{Putting}\text{value of k in equation}\left(\mathrm{i}\right),\text{we get}\\ (\text{5}+2\times -34)\text{x}+(-\text{3}+3\times -34)\text{y}-\text{1}-\text{23}\times -34=0\\ \Rightarrow -63\mathrm{x}-105\mathrm{y}-781=0\\ \Rightarrow 63\mathrm{x}+105\mathrm{y}-781=0\\ \mathrm{Therefore},\text{the required equation of the line is:}\\ 63\mathrm{x}+105\mathrm{y}-781=0.\end{array}$

Q.50 Find the new coordinates of the points in each of the following cases if the origin is shifted to the point (–3, –2) by a translation of axes.
(i) (1, 1) (ii) (0, 1) (iii) (5, 0) (iv) (–1, –2) (v) (3, –5)

Ans

(i) The coordinates of the new origin are h =– 3, k= –2, and the original coordinates are given to be x = 1, y = 1.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ =1 + 3 = 4 and y′ = 1 + 2 = 3
Thus, the coordinates of the point (1, 1) in the new system are (4, 3).

(ii) The coordinates of the new origin are h = – 3, k= –2, and the original coordinates are given to be x = 0, y = 1.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ = 0 + 3 = 3 and y′ = 1 + 2 = 3
Thus, the coordinates of the point (1, 1) in the new system are (3, 3).

(iii) The coordinates of the new origin are h =– 3,
k = –2, and the original coordinates are given to
be x = 5, y = 0.

The transformation relation between the old
coordinates (x, y) and the new coordinates (x′, y′) are given by x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have x′ =5 + 3 = 8 and y′ = 0 + 2 = 2
Thus, the coordinates of the point (5, 0) in the new system are (8, 2).

(iv) The coordinates of the new origin are h =– 3,
k = –2, and the original coordinates are given to be x = –1, y = – 2.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h
and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ = –1 + 3 = 2 and y′ = – 2 + 2 = 0
Thus, the coordinates of the point (–1, – 2) in the new system are (2, 0).

(v) The coordinates of the new origin are h =– 3, k= –2, and the original coordinates are given to be x = 3, y = – 5.
The transformation relation between the old coordinates (x, y) and the new coordinates (x′, y′) are given by
x = x′ + h i.e., x′ = x – h and y = y′ + k i.e., y′ = y – k
Substituting the values, we have
x′ = 3 + 3 = 6 and y′ = – 5 + 2 = – 3
Thus, the coordinates of the point (3, – 5) in the new system are (6, – 3).

Q.51 Find what the following equations become when the origin is shifted to the point (1, 1)
(i) x² + xy – 3y² – y + 2 = 0
(ii) xy – y² – x + y = 0
(iii) xy – x – y + 1 = 0

Ans

(i) Let coordinates of a point A changes from (x, y) to (x′, y′ ) in new coordinate axes whose origin has the coordinates
h = 1, k = 1. Therefore substituting
x = x′+1 and y = y′+1 in the given equation of the straight line, we get
(x′+1)² + (x′+1)(y′+1) – 3(y′+1)² – (y′+1)+ 2 = 0
or x’² + 2x’+ 1 + x’y’ + x’+ y’ + 1–3(y’² + 2y’+1)
– y’ – 1 + 2 = 0
x’² – 3y’² + x’y’ + 3 – 3y’² – 6y’ + 3 = 0
or x’² – 3y’² + x’y’ + 3x’ – 6y’ = 0
Therefore, the equation of the straight line in new system is x² – 3y² + xy + 3x – 6y = 0

(ii) Let coordinates of a point B changes from (x, y) to (x′, y′ ) in new coordinate axes whose origin has the coordinates
h = 1, k = 1. Therefore,
x = x′+1 and y = y′+1.Substituting values of x and y in the given equation of the straight line, we get
(x′+1) (y′+1) – (y′+1)² – (x′+1) + (y′+1) = 0
or x’y’ + x’ + y’ + 1 – y’² – 2y’ – 1 – x’ –1+ y’+1= 0
x’y’ – y’² = 0
Therefore, the equation of the straight line in new system is xy – y² = 0.

(iii) Let coordinates of a point B changes from (x, y) to (x′, y′ ) in new coordinate axes whose origin has the coordinates
h = 1, k = 1. Therefore, substituting
x = x′+1 and y = y′+1 in the given equation of the straight line, we get
(x′+1)(y′+1) – (x′+1) – (y′+1) + 1 = 0
or x’y’ + x’ + y’ + 1– x’ –1 – y’ – 1 + 1= 0
x’y’ = 0
Therefore, the equation of the straight line in new system is xy = 0.

Q.52 Find the values of k for which the line (k – 3) x – (4 – k²) y + k² –7k + 6 = 0 is
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

Ans

$\begin{array}{l}\mathrm{The}\text{given equation is:}\\ (\text{k}-\text{3})\text{x}-(\text{4}-{\text{k}}^{\text{2}})\text{y}+{\text{k}}^{\text{2}}-\text{7k}+\text{6}=0...\left(\mathrm{i}\right)\\ \left(\mathrm{a}\right)\mathrm{When}\text{line is parallel to x-axis,}\\ \text{Then x}=\text{0}\\ \Rightarrow \mathrm{k}-3=0\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \Rightarrow \mathrm{k}=3\end{array}$ $\begin{array}{l}\left(\mathrm{b}\right)\mathrm{When}\text{line is parallel to y-axis,}\\ \text{Then y}=\text{0}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4}-{\text{k}}^{\text{2}}=0\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \Rightarrow \mathrm{k}=\pm 2\\ \left(\mathrm{c}\right)\text{When given line passes through origin,}\\ \text{Putting x}=\text{0 and y}=\text{0 in equation}\left(\mathrm{i}\right),\text{we get}\\ (\text{k}-\text{3})\left(0\right)-(\text{4}-{\text{k}}^{\text{2}})\left(0\right)+{\text{k}}^{\text{2}}-\text{7k}+\text{6}=0\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em} k}}^{\text{2}}-\text{7k}+\text{6}=0\\ \Rightarrow (\mathrm{k}-6)(\mathrm{k}-1)=0\\ \Rightarrow \mathrm{k}=6,1\\ \mathrm{Thus},\text{value of k is 6 or 1.}\end{array}$

Q.53

$\begin{array}{l}\mathbf{\text{Find the values of θ and p, if the equation x cos θ + y sinθ = p is the normal form of the}}\\ \mathbf{\text{line}}\sqrt{\mathbf{\text{3}}}\mathbf{\text{x}}\mathbf{+}\mathbf{\text{y}}\mathbf{+}\mathbf{\text{2=0.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Equationofthegivenlineis}:\\ \sqrt{3}\mathrm{x}+\mathrm{y}+2=0...\left(\mathrm{i}\right)\\ \mathrm{Dividing}\text{equation}\left(\mathrm{i}\right)\text{by}\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=2,\text{we get}\\ -\frac{\sqrt{3}}{2}\mathrm{x}-\frac{1}{2}\mathrm{y}=\frac{2}{2}\\ \Rightarrow -\mathrm{xcos}\left(\frac{\mathrm{\pi }}{6}\right)-\mathrm{ysin}\left(\frac{\mathrm{\pi }}{6}\right)=1\\ \Rightarrow \mathrm{xcos}(\mathrm{\pi }+\frac{\mathrm{\pi }}{6})+\mathrm{ysin}(\mathrm{\pi }+\frac{\mathrm{\pi }}{6})=1\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{xcos}\left(\frac{7\mathrm{\pi }}{6}\right)+\mathrm{ysin}\left(\frac{7\mathrm{\pi }}{6}\right)=1\dots \left(\mathrm{ii}\right)\\ \mathrm{Since},\text{xcos}\mathrm{\theta }+\text{ysin}\mathrm{\theta }=\text{p is normal form of equation}\left(\mathrm{i}\right).\\ \mathrm{So},\text{comparing equation}\left(\mathrm{ii}\right)\text{and given normal form of}\\ \text{equation.We get}\\ \mathrm{\theta }=\frac{7\mathrm{\pi }}{6}\text{and p}=\text{1}\end{array}$

Q.54 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of line in intercepts form is:}\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1...\left(\mathrm{i}\right)\\ \mathrm{According}\text{to given conditions:}\\ \text{a}+\mathrm{b}=1...\left(\mathrm{ii}\right)\\ \mathrm{ab}=-6...\left(\mathrm{iii}\right)\\ \mathrm{Putting}\text{b}=\frac{-6}{\mathrm{a}}\text{in equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}+\frac{-6}{\mathrm{a}}=1\\ \Rightarrow \frac{{\mathrm{a}}^2-6}{\mathrm{a}}=1\\ \Rightarrow {\mathrm{a}}^2-6=\mathrm{a}\\ \Rightarrow {\mathrm{a}}^2-\mathrm{a}-6=0\\ (\mathrm{a}-3)(\mathrm{a}+2)=0\\ \Rightarrow \mathrm{a}=3,-2\end{array}$ $\begin{array}{l}\mathrm{and}\mathrm{b}=(1-3),(1+2)[âˆ\mu \mathrm{b}=1-\mathrm{a}]\\ \mathrm{b}=-2,3\\ \mathrm{Equation}\text{of line, when a}=\text{3 and b}=-\text{2:}\\ \frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{-2}=1\\ \frac{2\mathrm{x}-3\mathrm{y}}{6}=1\\ \Rightarrow 2\mathrm{x}-3\mathrm{y}=6\\ \mathrm{Equation}\text{of line, when a}=-\text{2 and b}=\text{3:}\\ \frac{\mathrm{x}}{-2}+\frac{\mathrm{y}}{3}=1\\ \frac{-3\mathrm{x}+2\mathrm{y}}{6}=1\\ \Rightarrow -3\mathrm{x}+2\mathrm{y}=6\end{array}$ $\begin{array}{l}\mathrm{}\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{the required equation of lines are:}\\ 2\mathrm{x}-3\mathrm{y}=6\text{and}-3\mathrm{x}+2\mathrm{y}=6.\end{array}$

Q.55

$\mathbf{\text{What are the points on the y}}\mathbf{–}\mathbf{\text{axis whose distance from the line}}\frac{\mathbf{\text{x}}}{\mathbf{\text{3}}}\mathbf{\text{+}}\frac{\mathbf{\text{y}}}{\mathbf{\text{4}}}\mathbf{\text{=1 is 4 units.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{the point on y-axis be P}(0,\mathrm{b})\text{and the given equation of line}\\ \text{is}\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{4}=1.\end{array}$ $\begin{array}{l}\mathrm{The}\text{distance of point P from line}=\left|\frac{\frac{0}{3}+\frac{\mathrm{b}}{4}-1}{\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{4}\right)}^2}}\right|\\ \Rightarrow 4=\left|\frac{\frac{0}{3}+\frac{\mathrm{b}}{4}-1}{\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{4}\right)}^2}}\right|\\ \Rightarrow 4=\left|\frac{\frac{\mathrm{b}}{4}-1}{\sqrt{\left(\frac{16+9}{9\times 16}\right)}}\right|\\ \Rightarrow 4=\left|\frac{\frac{\mathrm{b}}{4}-1}{\frac{5}{12}}\right|\\ \Rightarrow \pm 4\times \frac{5}{12}=\frac{\mathrm{b}}{4}-1\\ \mathrm{Taking}(+)\mathrm{ve}\text{sign:}\\ \frac{5}{3}=\frac{\mathrm{b}}{4}-1\\ \Rightarrow \frac{5}{3}+1=\frac{\mathrm{b}}{4}\\ \Rightarrow \frac{8}{3}=\frac{\mathrm{b}}{4}\\ \Rightarrow \mathrm{b}=\frac{32}{3}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{Therefore},\text{the point on y-axis is}(0,\frac{32}{3}).\\ \mathrm{Taking}(-)\mathrm{ve}\text{sign:}\\ -\frac{5}{3}=\frac{\mathrm{b}}{4}-1\\ \Rightarrow -\frac{5}{3}+1=\frac{\mathrm{b}}{4}\\ \Rightarrow -\frac{2}{3}=\frac{\mathrm{b}}{4}\\ \Rightarrow -\frac{8}{3}=\mathrm{b}\\ \mathrm{Therefore},\text{the point on y-axis is}(0,-\frac{8}{3}).\\ \mathrm{Thus},\text{the point on y-axis are}(0,-\frac{8}{3}),(0,\frac{32}{3}).\end{array}$

Q.56 Find perpendicular distance from the origin of the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).

Ans

Equation of the line joining two points is:

$\begin{array}{l}\mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}(\mathrm{x}-{\mathrm{x}}_1)\\ \mathrm{Equation}\text{of line joining}(\text{cos}\mathrm{\theta },\text{sin}\mathrm{\theta })\mathrm{}\text{and}(\text{cos}\mathrm{Ï•},\text{sin}\mathrm{Ï•})\text{is:}\\ \mathrm{y}-\text{sin}\mathrm{\theta }=\frac{\text{sin}\mathrm{Ï•}-\text{sin}\mathrm{\theta }}{\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta }}(\mathrm{x}-\text{cos}\mathrm{\theta })\\ (\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\mathrm{y}-\text{sin}\mathrm{\theta }(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })=(\text{sin}\mathrm{Ï•}-\text{sin}\mathrm{\theta })(\mathrm{x}-\text{cos}\mathrm{\theta })\\ (\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})\mathrm{x}+(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\mathrm{y}=\text{sin}\mathrm{\theta }(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\\ -\mathrm{cos\theta }(\text{sin}\mathrm{Ï•}-\text{sin}\mathrm{\theta })\end{array}$ $\begin{array}{l}(\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})\mathrm{x}+(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\mathrm{y}=\text{sin}\mathrm{\theta }\text{cos}\mathrm{Ï•}-\text{sin}\mathrm{\theta }\text{cos}\mathrm{\theta }-\text{cos}\mathrm{\theta }\text{sin}\mathrm{Ï•}\\ +\mathrm{cos\theta }\text{sin}\mathrm{\theta }\\ (\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})\mathrm{x}+(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\mathrm{y}=\text{sin}\mathrm{\theta }\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta }\text{sin}\mathrm{Ï•}\\ (\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})\mathrm{x}+(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\mathrm{y}=\text{sin}(\mathrm{\theta }-\mathrm{Ï•})\\ (\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})\mathrm{x}+(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\mathrm{y}-\text{sin}(\mathrm{\theta }-\mathrm{Ï•})=0\\ \text{Distance of equation}\left(\mathrm{i}\right)\text{from origin is:}\\ \text{d}=\left|\frac{(\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})\left(0\right)+(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })\left(0\right)-\text{sin}(\mathrm{\theta }-\mathrm{Ï•})}{\sqrt{{(\text{sin}\mathrm{\theta }-\text{sin}\mathrm{Ï•})}^2+{(\text{cos}\mathrm{Ï•}-\text{cos}\mathrm{\theta })}^2}}\right|\\ \Rightarrow \text{d}=\left|\frac{-\text{sin}(\mathrm{\theta }-\mathrm{Ï•})}{\sqrt{{\text{sin}}^{\text{2}}\mathrm{\theta }+{\text{sin}}^{\text{2}}\mathrm{Ï•}-2\text{sin}\mathrm{\theta }\text{sin}\mathrm{Ï•}+{\text{cos}}^{\text{2}}\mathrm{Ï•}+{\text{cos}}^{\text{2}}\mathrm{\theta }-2\text{cos}\mathrm{Ï•}\text{cos}\mathrm{\theta }}}\right|\\ \Rightarrow \text{d}=\left|\frac{-\text{sin}(\mathrm{\theta }-\mathrm{Ï•})}{\sqrt{{\text{sin}}^{\text{2}}\mathrm{\theta }+{\text{cos}}^{\text{2}}\mathrm{\theta }-2\text{sin}\mathrm{\theta }\text{sin}\mathrm{Ï•}+{\text{cos}}^{\text{2}}\mathrm{Ï•}+{\text{sin}}^{\text{2}}\mathrm{Ï•}-2\text{cos}\mathrm{Ï•}\text{cos}\mathrm{\theta }}}\right|\\ \Rightarrow \text{d}=\left|\frac{-\text{sin}(\mathrm{\theta }-\mathrm{Ï•})}{\sqrt{1-2\text{sin}\mathrm{\theta }\text{sin}\mathrm{Ï•}+1-2\text{cos}\mathrm{Ï•}\text{cos}\mathrm{\theta }}}\right|\\ \Rightarrow \text{d}=\left|\frac{-\text{sin}(\mathrm{\theta }-\mathrm{Ï•})}{\sqrt{2-2(\text{sin}\mathrm{\theta }\text{sin}\mathrm{Ï•}+\text{cos}\mathrm{Ï•}\text{cos}\mathrm{\theta })}}\right|\\ \Rightarrow \text{d}=\left|\frac{\text{sin}(\mathrm{Ï•}-\mathrm{\theta })}{\sqrt{2\{1-\text{cos}(\mathrm{Ï•}-\mathrm{\theta })\}}}\right|\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow \text{d}=\left|\frac{\text{sin}(\mathrm{Ï•}-\mathrm{\theta })}{\sqrt{2\left\{2{\sin }^2\left(\frac{\mathrm{Ï•}-\mathrm{\theta }}{2}\right)\right\}}}\right|\\ \Rightarrow \text{d}=\left|\frac{\text{sin}(\mathrm{Ï•}-\mathrm{\theta })}{\sqrt{4{\sin }^2\left(\frac{\mathrm{Ï•}-\mathrm{\theta }}{2}\right)}}\right|\\ \Rightarrow \text{d}=\left|\frac{\text{sin}(\mathrm{Ï•}-\mathrm{\theta })}{2\sin \left(\frac{\mathrm{Ï•}-\mathrm{\theta }}{2}\right)}\right|\\ \Rightarrow \text{d}=\frac{1}{2}\frac{\left|\text{sin}(\mathrm{Ï•}-\mathrm{\theta })\right|}{\left|\sin \left(\frac{\mathrm{Ï•}-\mathrm{\theta }}{2}\right)\right|}\\ \mathrm{Thus},\text{the perpendicular distance from the origin of the line}\\ \text{joining the points}(\text{cos}\mathrm{\theta },\text{sin}\mathrm{\theta })\mathrm{}\mathrm{a}\text{nd}(\text{cos}\mathrm{Ï•},\text{sin}\mathrm{Ï•})\text{is}\frac{1}{2}\frac{\left|\text{sin}(\mathrm{Ï•}-\mathrm{\theta })\right|}{\left|\sin \left(\frac{\mathrm{Ï•}-\mathrm{\theta }}{2}\right)\right|}.\end{array}$

Q.57 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Ans

$\begin{array}{l}\mathrm{The}\text{given lines are:}\\ \text{x}-\text{7y}+\text{5}=0...\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}+\text{y}=0...\left(\mathrm{ii}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{x}=-\frac{5}{22}\text{and y}=\frac{15}{22}\\ \mathrm{Slope}\text{of line parallel to y-axis}=\tan 90\mathrm{°}\\ \mathrm{Equation}\text{of line parallel to y-axis and passing through}\\ \text{the point}(-\frac{5}{22},\frac{15}{22})\text{is}\\ \text{y}-\frac{15}{22}=\tan 90\mathrm{°}(\mathrm{x}+\frac{5}{22})\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}y}-\frac{15}{22}=\frac{1}{0}(\mathrm{x}+\frac{5}{22})\\ \Rightarrow 0=(\mathrm{x}+\frac{5}{22})\\ \Rightarrow \mathrm{x}=-\frac{5}{22}\\ \mathrm{Thus},\text{the equation of required line is}\mathrm{x}=-\frac{5}{22}.\end{array}$

Q.58

$\begin{array}{l}\mathbf{\text{Find the equation of a line drawn perpendicular to the line}}\frac{\mathbf{\text{x}}}{\mathbf{\text{4}}}\mathbf{\text{+}}\frac{\mathbf{\text{y}}}{\mathbf{\text{6}}}\mathbf{\text{=1\hspace{0.17em}through the point,}}\\ \mathbf{\text{where it meets the y-axis.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Equation}\text{of given line is:}\frac{\text{x}}{4}+\frac{\text{y}}{6}=1...\left(\text{i}\right)\\ \mathrm{Since},\text{line meets at y-axis.}\\ \therefore \text{Putting x}=0\text{in equation}\left(\text{i}\right),\text{we get}\\ \frac{0}{4}+\frac{\text{y}}{6}=1\\ \Rightarrow \text{y}=6\\ \mathrm{Coordinate}\text{s of the point where given line meets aty-axis is}\left(0,6\right).\\ \mathrm{Slope}\text{of given line}=-\frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{6}\right)}\left[âˆ\mu \text{m}=-\frac{\text{A}}{\text{B}}\right]\\ =-\frac{3}{2}\\ \mathrm{Slope}\text{of perpendicular}=-\frac{1}{\left(-\frac{3}{2}\right)}\\ =\frac{2}{3}\\ \mathrm{Equation}\text{of perpendicular to the given line is:}\\ \text{y}-6=\frac{2}{3}\left(\text{x}-0\right)\\ \Rightarrow 3\left(\text{y}-6\right)=2\text{x}\\ \Rightarrow 3\text{y}-18=2\text{x}\\ \Rightarrow 2\text{x}-3\text{y}+18=0\\ \mathrm{Therefore},2\text{x}-3\text{y}+18=0\text{is the required equation ofperpendicular line.}\end{array}$

Q.59 Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Ans

$\begin{array}{l}\mathrm{Given}\text{equations of sides of triangle are:}\\ \text{y}-\mathrm{x}=\text{0}...\left(\mathrm{i}\right)\\ \text{x}+\mathrm{y}=\text{0}...\left(\mathrm{ii}\right)\\ \text{x}-\mathrm{k}=\text{0}...\left(\mathrm{iii}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{x}=0\text{, y}=0\\ \mathrm{Solving}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{x}=\mathrm{k}\text{, y}=-\mathrm{k}\end{array}$

$\begin{array}{l}\mathrm{Solving}\text{equation}\left(\mathrm{iii}\right)\text{and equation}\left(\mathrm{i}\right),\text{we have}\\ \text{x}=\mathrm{k}\text{, y}=\mathrm{k}\\ \mathrm{Therefore},\text{the coordinates of vertices of triangle are:}\\ \text{A}(0,0),\mathrm{B}(\mathrm{k},-\mathrm{k})\text{and C}(\mathrm{k},\mathrm{k}).\\ \mathrm{Area}\left(\mathrm{OPQ}\right)=\frac{1}{2}\times \mathrm{PQ}\times \mathrm{OR}\\ =\frac{1}{2}\times 2\mathrm{k}\times \mathrm{k}\\ ={\mathrm{k}}^2\text{square units}\\ {\text{Thus, the area of triangle is k}}^{\text{2}}\text{square units.}\end{array}$

Q.60 Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Ans

$\begin{array}{l}\mathrm{Given}\text{equations are:}\\ \text{3x}+\text{y}-\text{2}=0...\left(\mathrm{i}\right)\\ \text{px}+\text{2y}-\text{3}=0...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}2x}-\text{y}-\text{3}=0...\left(\mathrm{iii}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{x}=1\text{and y}=-1\\ \mathrm{Putting}\text{values of x and y in equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{p}\left(1\right)+\text{2}(-1)-\text{3}=0\\ \Rightarrow \mathrm{p}-2-3=0\\ \Rightarrow \mathrm{p}=5\\ \mathrm{Thus},\text{the value of p is 5.}\end{array}$

Q.61 If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃(c₁– c₂) = 0.

Ans

$\begin{array}{l}\mathrm{The}\text{given equations are:}\\ \text{y}={\text{m}}_{\text{1}}\text{x}+{\text{c}}_{\text{1}}...\left(\mathrm{i}\right)\\ \text{y}={\text{m}}_{\text{2}}\text{x}+{\text{c}}_{\text{2}}...\left(\mathrm{ii}\right)\\ \text{y}={\text{m}}_{\text{3}}\text{x}+{\text{c}}_{\text{3}}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ {\text{\hspace{0.17em}\hspace{0.17em} m}}_{\text{1}}\text{x}+{\text{c}}_{\text{1}}={\text{m}}_{\text{2}}\text{x}+{\text{c}}_{\text{2}}\\ \Rightarrow ({\text{m}}_{\text{1}}-{\text{m}}_{\text{2}})\text{x}={\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}\\ \Rightarrow \mathrm{x}=\frac{{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}}{{\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}}\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right),\text{we have}\\ \text{y}={\text{m}}_{\text{1}}\left(\frac{{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}}{{\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}}\right)+{\text{c}}_{\text{1}}\\ \text{y}=\frac{{\text{m}}_{\text{1}}{\text{c}}_{\text{2}}-{\text{m}}_{\text{1}}{\text{c}}_{\text{1}}+{\text{c}}_{\text{1}}{\text{m}}_{\text{1}}-{\text{c}}_{\text{1}}{\text{m}}_{\text{2}}}{{\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}}\\ \text{y}=\frac{{\text{m}}_{\text{1}}{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}{\text{m}}_{\text{2}}}{{\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}}\\ \mathrm{Since},\text{given three equations are concurrent, so values}\\ \text{of x and y will satisfy equation}\left(\mathrm{iii}\right),\text{then}\\ \frac{{\text{m}}_{\text{1}}{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}{\text{m}}_{\text{2}}}{{\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}}={\text{m}}_{\text{3}}\frac{{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}}{{\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}}+{\text{c}}_{\text{3}}\end{array}$ $\begin{array}{l}\Rightarrow {\text{m}}_{\text{1}}{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}{\text{m}}_{\text{2}}={\text{m}}_{\text{3}}{\text{c}}_{\text{2}}-{\text{m}}_{\text{3}}{\text{c}}_{\text{1}}+{\text{c}}_{\text{3}}{\text{m}}_{\text{1}}-{\text{c}}_{\text{3}}{\text{m}}_{\text{2}}\\ \Rightarrow {\text{m}}_{\text{1}}{\text{c}}_{\text{2}}-{\text{c}}_{\text{1}}{\text{m}}_{\text{2}}-{\text{m}}_{\text{3}}{\text{c}}_{\text{2}}+{\text{m}}_{\text{3}}{\text{c}}_{\text{1}}-{\text{c}}_{\text{3}}{\text{m}}_{\text{1}}+{\text{c}}_{\text{3}}{\text{m}}_{\text{2}}=0\\ \Rightarrow {\text{m}}_{\text{1}}({\text{c}}_{\text{2}}-{\text{c}}_{\text{3}})+{\mathrm{m}}_2({\mathrm{c}}_3-{\mathrm{c}}_1)+{\mathrm{m}}_3({\mathrm{c}}_1-{\mathrm{c}}_2)=0\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\text{proved.}\end{array}$

Q.62 Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.

Ans

$\begin{array}{l}\mathrm{The}\text{given line is x}-\text{2y}=\text{3.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}The slope}\left({\mathrm{m}}_1\right)\text{of given line}=\frac{1}{2}[âˆ\mu \mathrm{m}=-\frac{\mathrm{A}}{\mathrm{B}}]\\ \mathrm{Angle}\text{between two lines,}\mathrm{\theta }=45\mathrm{°}\\ \mathrm{Let}\text{s}\mathrm{lope}\mathrm{of}\text{t}\mathrm{he}{\text{other line,m}}_{\text{2}}=\mathrm{m}\\ \mathrm{Since},\text{tan}\mathrm{\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ \Rightarrow \tan 45\mathrm{°}=\left|\frac{\mathrm{m}-\frac{1}{2}}{1+\left(\frac{1}{2}\right)\mathrm{m}}\right|\\ \Rightarrow \pm 1=\frac{2\mathrm{m}-1}{2+\mathrm{m}}\\ \Rightarrow \pm (2+\mathrm{m})=2\mathrm{m}-1\\ \Rightarrow \mathrm{m}=3,-\frac{1}{3}\\ \mathrm{So},\text{the equation of line having slope 3 and passing}\\ \text{through the point}(3,2)\text{is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}-\text{2}=3(\mathrm{x}-3)\end{array}$ $\begin{array}{l}\Rightarrow 3\mathrm{x}-\mathrm{y}=9-2\\ \Rightarrow 3\mathrm{x}-\mathrm{y}=9-2\\ \Rightarrow 3\mathrm{x}-\mathrm{y}=7\\ \mathrm{So},\text{the equation of line having slope}-\frac{1}{3}\text{and passing}\\ \text{through the point}(3,2)\text{is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}-\text{2}=-\frac{1}{3}(\mathrm{x}-3)\\ \Rightarrow 3\mathrm{y}-6=-\mathrm{x}+3\\ \Rightarrow \mathrm{x}+3\mathrm{y}=6+3\\ \Rightarrow \mathrm{x}+3\mathrm{y}=9\\ \mathrm{Thus},\text{the required equation of line is x}+3\text{y}=\text{9 or x}+\text{3y}=\text{9.}\end{array}$

Q.63 Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Ans

$\begin{array}{l}\text{Equation of the given lines}\mathrm{are}:\\ \text{4x}+\text{7y}-\text{3}=0...\left(\text{i}\right)\\ \text{and 2x}-\text{3y}+\text{1}=0...\left(\mathrm{ii}\right)\\ \mathrm{Solving}\text{equation}\left(\text{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{x}=\frac{1}{13}\text{and y}=\frac{5}{13}\\ \mathrm{Equation}\text{of line having equal intercepts on the axis;}\\ \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}=1...\left(\mathrm{iii}\right)\\ \mathrm{Since},\text{equation}\left(\mathrm{iii}\right)\text{is passing through}\left(\frac{1}{13},\frac{5}{13}\right),\mathrm{so}\\ \frac{\left(\frac{1}{13}\right)}{\text{a}}+\frac{\left(\frac{5}{13}\right)}{\text{a}}=1\\ \Rightarrow \frac{1}{\text{a}}+\frac{5}{\text{a}}=13\\ \Rightarrow \frac{6}{\text{a}}=13\\ \Rightarrow \text{a}=\frac{6}{13}\\ \mathrm{Thus},\text{the required equation of the line is}\\ \frac{\text{x}}{\left(\frac{6}{13}\right)}+\frac{\text{y}}{\left(\frac{6}{13}\right)}=1\left[\mathrm{From}\text{equation}\left(\mathrm{iii}\right)\right]\\ \Rightarrow \text{x}+\text{y}=\frac{6}{13}\\ \Rightarrow 13\text{x}+13\text{y}=6\\ \mathrm{Thus},\text{the required equation of the line is}13\text{x}+13\text{y}=6.\end{array}$

Q.64

$\begin{array}{l}\mathbf{\text{Show that the equation of the line passing through theorigin and making an angleθ with the line}}\\ \mathbf{\text{y}}\mathbf{=}\mathbf{\text{mx}}\mathbf{+}\mathbf{\text{c is}}\frac{\mathbf{\text{y}}}{\mathbf{\text{x}}}\mathbf{\text{=}}\frac{\mathbf{\text{m}}\mathbf{\pm }\mathbf{\text{tanθ}}}{\mathbf{\text{1}}\mathbf{∓}\mathbf{\text{mtanθ}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Slope}\text{of given line}(\mathrm{y}=\mathrm{mx}+\mathrm{c}){\text{, m}}_{\text{1}}=\mathrm{m}\\ \mathrm{Let}\text{slope of line making an angle}\mathrm{\theta }\mathrm{with}\mathrm{the}\mathrm{given}\mathrm{line}={\mathrm{m}}_2\end{array}$ $\begin{array}{l}\mathrm{Then},\text{tan}\mathrm{\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ \Rightarrow \text{tan}\mathrm{\theta }=\left|\frac{{\mathrm{m}}_2-\mathrm{m}}{1+{\mathrm{mm}}_2}\right|\\ \Rightarrow \pm \text{\hspace{0.17em}tan}\mathrm{\theta }=\frac{{\mathrm{m}}_2-\mathrm{m}}{1+{\mathrm{mm}}_2}\\ \Rightarrow \pm \text{\hspace{0.17em}tan}\mathrm{\theta }(1+{\mathrm{mm}}_2)={\mathrm{m}}_2-\mathrm{m}\\ \pm \text{\hspace{0.17em}tan}\mathrm{\theta }\pm \text{\hspace{0.17em}tan}\mathrm{\theta }.{\mathrm{mm}}_2={\mathrm{m}}_2-\mathrm{m}\\ \Rightarrow \pm \text{\hspace{0.17em}tan}\mathrm{\theta }+\mathrm{m}={\mathrm{m}}_2∓\text{\hspace{0.17em}tan}\mathrm{\theta }.{\mathrm{mm}}_2\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow \pm \text{\hspace{0.17em}tan}\mathrm{\theta }+\mathrm{m}={\mathrm{m}}_2(1∓\mathrm{m}\text{tan}\mathrm{\theta })\\ \Rightarrow {\mathrm{m}}_2=\frac{\mathrm{m}\pm \text{\hspace{0.17em}tan}\mathrm{\theta }}{1∓\mathrm{m}\text{tan}\mathrm{\theta }}\\ \mathrm{Equation}{\text{of line with slope m}}_{\text{1}}\text{and passing}\\ \text{through origin}(0,0)\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}-\text{0}=\frac{\mathrm{m}\pm \text{\hspace{0.17em}tan}\mathrm{\theta }}{1∓\mathrm{m}\text{tan}\mathrm{\theta }}(\mathrm{x}-0)\\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{m}\pm \text{\hspace{0.17em}tan}\mathrm{\theta }}{1∓\mathrm{m}\text{tan}\mathrm{\theta }}\end{array}$

Q.65 In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Ans

$\mathrm{Equation}\text{of line joining the points}(-1,1)\text{and}(5,7)\text{is}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} y}-1=\frac{7-1}{5+1}(\mathrm{x}+1)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}-1=\frac{6}{6}(\mathrm{x}+1)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}y}-1=\mathrm{x}+1\\ \Rightarrow \mathrm{x}-\mathrm{y}+2=0...\left(\mathrm{i}\right)\\ \mathrm{Given}\text{equation is:}\\ \text{\hspace{0.17em}\hspace{0.17em}x}+\text{y}=\text{4}...\left(\mathrm{ii}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{x}=\text{1 and y}=3\\ \mathrm{So},\text{the intersection point of both the lines is}(1,3).\\ \mathrm{Let}\text{point}(1,3)\text{divides the line segment joining points}\\ (-1,1)\text{and}(5,7)\text{in the ratio of m:1, then}\\ (1,3)=\{\frac{\mathrm{m}\left(5\right)+1(-1)}{\mathrm{m}+1},\frac{\mathrm{m}\left(7\right)+1\left(1\right)}{\mathrm{m}+1}\}\\ \Rightarrow (1,3)=\{\frac{5\mathrm{m}-1}{\mathrm{m}+1},\frac{7\mathrm{m}+1}{\mathrm{m}+1}\}\\ \Rightarrow 1=\frac{5\mathrm{m}-1}{\mathrm{m}+1}\text{and 3}=\frac{7\mathrm{m}+1}{\mathrm{m}+1}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{m}+1=5\mathrm{m}-1\text{and}3\mathrm{m}+3=7\mathrm{m}+1\\ \Rightarrow 4\mathrm{m}=2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and 4m}=2\\ \Rightarrow \mathrm{m}=\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}m}=\frac{1}{2}\\ \mathrm{Thus},\text{the required ratio is 1:}2.\end{array}$

Q.66 Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Ans

$\begin{array}{l}\text{The given lines are:}\\ \text{4x}+\text{7y}+\text{5}=0...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em} 2x}-\text{y}=0...\left(\mathrm{ii}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{x}=-\frac{5}{18},\text{y}=-\frac{5}{9}\\ \mathrm{Since},\\ \text{Distance of point}(1,2)\text{from equation}\left(\mathrm{i}\right)=\text{distance between}\\ \text{intersection point}(-\frac{5}{18},-\frac{5}{9})\text{\hspace{0.17em}\hspace{0.17em}and the point}(1,2)\\ =\sqrt{{(1+\frac{5}{18})}^2+{(2+\frac{5}{9})}^2}\\ =\sqrt{{\left(\frac{23}{18}\right)}^2+{\left(\frac{23}{9}\right)}^2}\\ =\frac{23}{9}\sqrt{\frac{1}{4}+1}\\ =\frac{23\sqrt{5}}{9\times 2}\\ =\frac{23\sqrt{5}}{18}\mathrm{units}\end{array}$ $\mathrm{Thus},\text{the distance of point}(1,2)\text{from line}\left(\mathrm{i}\right)\text{along line}\left(\mathrm{ii}\right)$ $\text{is}\frac{23\sqrt{5}}{18}\mathrm{units}.$

Q.67 Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Ans

$\begin{array}{l}\mathrm{Let}\text{a straight line passing through}(-1,2)\text{may intersect line}\\ \text{x}+\text{y}=\text{4 at}(\mathrm{h},\mathrm{k})\text{.}\\ \text{Since, point}(\mathrm{h},\mathrm{k})\text{lies on line x}+\text{y}=\text{4. Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} h}+\mathrm{k}=4\Rightarrow \mathrm{k}=4-\mathrm{h}...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{distance between}(-1,2)\text{and}(\mathrm{h},\mathrm{k})=3\\ \sqrt{{(\mathrm{h}+1)}^2+{(\mathrm{k}-2)}^2}=3\\ {\mathrm{h}}^2+2\mathrm{h}+1+{\mathrm{k}}^2-4\mathrm{k}+4=9\\ {\mathrm{h}}^2+2\mathrm{h}+{\mathrm{k}}^2-4\mathrm{k}=4\\ \mathrm{Putting}\text{value of k in equation}\left(\mathrm{i}\right),\text{we get}\\ {\mathrm{h}}^2+2\mathrm{h}+{(4-\mathrm{h})}^2-4(4-\mathrm{h})=4\\ {\mathrm{h}}^2+2\mathrm{h}+16-8\mathrm{h}+{\mathrm{h}}^2-16+4\mathrm{h}=4\\ \Rightarrow {\mathrm{h}}^2-\mathrm{h}-2=0\\ \Rightarrow (\mathrm{h}-2)(\mathrm{h}+1)=0\\ \Rightarrow \mathrm{h}=2,-1\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}k}=(4-2),(4+1)\\ =2,5\\ \mathrm{So},\text{the possible intersection points are}(2,2)\text{and}(-1,5).\\ \mathrm{Slope}\text{of line passing through}(-1,2)\end{array}$ $\begin{array}{l}\text{and}(2,2)=\frac{2-2}{2+1}\\ \mathrm{m}=0\\ \Rightarrow \mathrm{Line}\text{passing through}(-1,2)\text{is parallel to x-axis.}\\ \mathrm{Slope}\text{of line passing through}(-1,2)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and}(-1,5)=\frac{5-2}{-1+1}\\ \mathrm{m}=\frac{3}{0}=\mathrm{\infty }\\ \Rightarrow \mathrm{Line}\text{passing through}(-1,2)\text{is parallel to y-axis.}\\ \text{Therefore, line is parallel to x-axis or y-axis.}\end{array}$

Q.68 The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Ans

$\begin{array}{l}\mathrm{End}\text{points of hypotenuse}=(1,3)\text{and}(-4,1)\\ \mathrm{Let}\text{intersection points of legs}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of triangle}=(\mathrm{h},\mathrm{k})\\ \mathrm{Slope}\left({\mathrm{m}}_1\right)\text{of side having end points}\\ (\mathrm{h},\mathrm{k})\mathrm{and}(1,3)=\frac{3-\mathrm{k}}{1-\mathrm{h}}\\ \mathrm{Slope}\left({\mathrm{m}}_2\right)\text{of side having end points}\\ (\mathrm{h},\mathrm{k})\mathrm{and}(-4,1)=\frac{1-\mathrm{k}}{-4-\mathrm{h}}\end{array}$ $\begin{array}{l}\mathrm{Since},\text{both sides are perpendicular.}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_1{\mathrm{m}}_2=-1\\ \Rightarrow \frac{3-\mathrm{k}}{1-\mathrm{h}}\times \frac{1-\mathrm{k}}{-4-\mathrm{h}}=-1\\ \Rightarrow \frac{3-\mathrm{k}}{1-\mathrm{h}}\times \frac{1-\mathrm{k}}{4+\mathrm{h}}=1\\ \Rightarrow 3-4\mathrm{k}+{\mathrm{k}}^2=4-3\mathrm{h}-{\mathrm{h}}^2\\ \Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2+3\mathrm{h}-4\mathrm{k}=1...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{equation}\left(\mathrm{i}\right)\text{is quadratic and other condition is not}\\ \text{given, so value of h and k can not be found.}\\ \text{There may be many solutions for different values of h and k.}\end{array}$

Q.69 Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Ans

$\begin{array}{l}\mathrm{Let}\text{image of point}(3,8)\text{is}(\mathrm{h},\mathrm{k}).\\ \mathrm{Since},\text{point and its image is always equidistant from mirror.}\\ \text{So, the mid-point of}(3,8)\text{and}(\mathrm{h},\mathrm{k})=(\frac{3+\mathrm{h}}{2},\frac{8+\mathrm{k}}{2})\\ \mathrm{Since},\text{line x}+\text{3y}=\text{7 is assumed as a mirror, so mid-point}\\ \text{will lie on it.}\\ \therefore \frac{3+\mathrm{h}}{2}+\text{3}\left(\frac{8+\mathrm{k}}{2}\right)=\text{7}\\ \Rightarrow 3+\mathrm{h}+24+3\mathrm{k}=14\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{h}+3\mathrm{k}=-13\dots \left(\mathrm{i}\right)\\ \mathrm{Slope}\text{of}(3,8)\text{and}(\mathrm{h},\mathrm{k})=\frac{\mathrm{k}-8}{\mathrm{h}-3}\\ \mathrm{Slope}\text{of line}(\text{x}+\text{3y}=\text{7})=-\frac{1}{3}\\ \mathrm{Since},\text{line segment joining the points}(3,8)\text{and}(\mathrm{h},\mathrm{k})\text{is}\\ \text{perpendicular to line x}+\text{3y}=\text{7}.\\ \mathrm{So},\frac{\mathrm{k}-8}{\mathrm{h}-3}\times -\frac{1}{3}=-1\\ \Rightarrow \frac{\mathrm{k}-8}{\mathrm{h}-3}\times \frac{1}{3}=1\\ \Rightarrow \mathrm{k}-8=3\mathrm{h}-9\\ \Rightarrow 3\mathrm{h}-\mathrm{k}=1...\left(\mathrm{ii}\right)\\ \mathrm{Solving}\text{equation\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{h}=-1\text{and k}=-4\\ \mathrm{Therefore},\text{the image of point}(3,8)\text{w.r.t. line}\\ \text{x}+\text{3y}=\text{7 is}(-1,-4).\end{array}$

Q.70 If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

Ans

$\begin{array}{l}\mathrm{The}\text{equations of given lines are:}\\ \text{y}=\text{3x}+\text{1}...\left(\mathrm{i}\right)\\ \text{2y}=\text{x}+\text{3}\\ \text{\hspace{0.17em}y}=\frac{1}{2}\text{x}+\text{3}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}y}=\text{mx}+\text{4}...\left(\mathrm{iii}\right)\\ \mathrm{Slope}\left({\mathrm{m}}_1\right)\text{of line}\left(\mathrm{i}\right)=3\\ \mathrm{Slope}\left({\mathrm{m}}_2\right)\text{of line}\left(\mathrm{ii}\right)=\frac{1}{2}\\ \mathrm{Slope}\text{of line}\left(\mathrm{iii}\right)=\mathrm{m}\\ \mathrm{Angle}\text{between line}\left(\mathrm{i}\right)\text{and line}\left(\mathrm{iii}\right)=\mathrm{Angle}\text{between line}\left(\mathrm{ii}\right)\\ \text{and line}\left(\mathrm{iii}\right)\\ \left|\frac{\mathrm{m}-3}{1+3\mathrm{m}}\right|=\left|\frac{\mathrm{m}-\frac{1}{2}}{1+\frac{1}{2}\mathrm{m}}\right|\\ \Rightarrow \left|\frac{\mathrm{m}-3}{1+3\mathrm{m}}\right|=\left|\frac{2\mathrm{m}-1}{2+\mathrm{m}}\right|\\ \Rightarrow \frac{\mathrm{m}-3}{1+3\mathrm{m}}=\pm \frac{2\mathrm{m}-1}{2+\mathrm{m}}\\ \mathrm{Case}1:\text{If \hspace{0.17em}}\frac{\mathrm{m}-3}{1+3\mathrm{m}}=\frac{2\mathrm{m}-1}{2+\mathrm{m}}\\ \Rightarrow (\mathrm{m}-3)(2+\mathrm{m})=(2\mathrm{m}-1)(1+3\mathrm{m})\\ \Rightarrow {\mathrm{m}}^2-\mathrm{m}-6=6{\mathrm{m}}^2-\mathrm{m}-1\\ \Rightarrow 0=5{\mathrm{m}}^2+5\\ \Rightarrow {\mathrm{m}}^2+1=0\\ \Rightarrow \mathrm{m}=\sqrt{-1},\text{not real.}\\ \mathrm{Case}2:\text{If \hspace{0.17em}}\frac{\mathrm{m}-3}{1+3\mathrm{m}}=-\frac{2\mathrm{m}-1}{2+\mathrm{m}}\end{array}$ $\begin{array}{l}\Rightarrow (\mathrm{m}-3)(2+\mathrm{m})=-(2\mathrm{m}-1)(1+3\mathrm{m})\\ \Rightarrow {\mathrm{m}}^2-\mathrm{m}-6=-6{\mathrm{m}}^2+\mathrm{m}+1\\ \Rightarrow 7{\mathrm{m}}^2-2\mathrm{m}-7=0\\ \mathrm{By}\text{quadratic formula:}\\ \text{m}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ =\frac{-(-2)\pm \sqrt{{(-2)}^2-4\left(7\right)(-7)}}{2\left(7\right)}\\ =\frac{2\pm \sqrt{4+196}}{14}\\ \text{m}=\frac{1\pm 5\sqrt{2}}{7}\\ \mathrm{Therefore}\text{, the values of m are}\frac{1\pm 5\sqrt{2}}{7}.\end{array}$

Q.71 If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Ans

$\begin{array}{l}\text{If sum of the perpendicular distances of avariable point P}\left(\text{x},\text{y}\right)\text{from the lines}\\ \text{x}+\text{y}–\text{5}=0\text{and 3x}–\text{2y}+\text{7}=0\text{is always 1}0.\\ \text{Show that P must move on a line}.\\ \mathrm{Given}\text{lines are:}\\ \text{x}+\text{y}-\text{5}=0...\left(\text{i}\right)\end{array}$ $\begin{array}{l}\text{3x}-\text{2y}+\text{7}=0...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}Distance of P}(\mathrm{x},\mathrm{y})\text{from line}\left(\mathrm{i}\right)=\left|\frac{\text{x}+\text{y}-\text{5}}{\sqrt{1^2+1^2}}\right|\\ =\left|\frac{\text{x}+\text{y}-\text{5}}{\sqrt{2}}\right|\\ \text{Distance of P}(\mathrm{x},\mathrm{y})\text{from line}\left(\mathrm{ii}\right)=\left|\frac{\text{3x}-\text{2y}+\text{7}}{\sqrt{3^2+{(-2)}^2}}\right|\\ =\left|\frac{\text{3x}-\text{2y}+\text{7}}{\sqrt{9+4}}\right|\\ =\left|\frac{\text{3x}-\text{2y}+\text{7}}{\sqrt{13}}\right|\\ \mathrm{According}\text{to given condition:}\\ \frac{|\text{x}+\text{y}-\text{5}|}{\sqrt{2}}+\frac{|\text{3x}-\text{2y}+\text{7}|}{\sqrt{13}}=10\\ \sqrt{13}(\text{x}+\text{y}-\text{5})\pm \sqrt{2}(\text{3x}-\text{2y}+\text{7})=10\sqrt{26}\\ \mathrm{Case}\text{1:}\\ \sqrt{13}(\text{x}+\text{y}-\text{5})+\sqrt{2}(\text{3x}-\text{2y}+\text{7})=10\sqrt{26}\\ \sqrt{13}\mathrm{x}+\sqrt{13}\mathrm{y}-5\sqrt{13}+3\sqrt{2}\mathrm{x}-2\sqrt{2}\mathrm{y}+7\sqrt{2}=10\sqrt{26}\\ \Rightarrow \sqrt{13}\mathrm{x}+\sqrt{13}\mathrm{y}+3\sqrt{2}\mathrm{x}-2\sqrt{2}\mathrm{y}=10\sqrt{26}+5\sqrt{13}-7\sqrt{2}\\ \mathrm{So},\mathrm{locus}\text{of point P}(\mathrm{x},\mathrm{y})\text{is a linear equation.}\\ \mathrm{Case}\text{2:}\\ \sqrt{13}(\text{x}+\text{y}-\text{5})-\sqrt{2}(\text{3x}-\text{2y}+\text{7})=10\sqrt{26}\end{array}$ $\begin{array}{l}\sqrt{13}\mathrm{x}+\sqrt{13}\mathrm{y}-5\sqrt{13}-3\sqrt{2}\mathrm{x}+2\sqrt{2}\mathrm{y}-7\sqrt{2}=10\sqrt{26}\\ \Rightarrow \sqrt{13}\mathrm{x}+\sqrt{13}\mathrm{y}-3\sqrt{2}\mathrm{x}+2\sqrt{2}\mathrm{y}=10\sqrt{26}+5\sqrt{13}+7\sqrt{2}\\ \mathrm{So},\mathrm{locus}\text{of point P}(\mathrm{x},\mathrm{y})\text{is a linear equation.}\end{array}$

Q.72 Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Ans

$\begin{array}{l}\mathrm{Given}\text{parallel lines are:}\\ \text{9x}+\text{6y}-\text{7}=0...\left(\mathrm{i}\right)\\ \mathrm{slope}\left({\mathrm{m}}_1\right)=-\frac{9}{6}\\ =-\frac{3}{2}\\ \text{and}\\ \text{3x}+\text{2y}+\text{6}=0...\left(\mathrm{ii}\right)\\ \mathrm{slope}\left({\mathrm{m}}_2\right)=-\frac{3}{2}\\ \mathrm{Since},\text{line equidistant from line}\left(\mathrm{i}\right)\text{and line}\left(\mathrm{ii}\right)\text{is also}\\ \text{parallel to given lines.}\\ \text{Let any point on the line be P}(\mathrm{h},\mathrm{k}),\text{then}\\ \text{Distance of line}\left(\mathrm{i}\right)\text{from the point P}=\text{Distance of line}\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}from the point P}\\ \Rightarrow \frac{|\text{9h}+\text{6k}-\text{7}|}{\sqrt{9^2+6^2}}=\frac{|\text{3h}+\text{2k}+\text{6}|}{\sqrt{3^2+2^2}}\\ \Rightarrow \frac{(\text{9h}+\text{6k}-\text{7})}{\sqrt{117}}=\pm \frac{(\text{3h}+\text{2k}+\text{6})}{\sqrt{13}}\end{array}$ $\begin{array}{l}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}9h}+\text{6k}-\text{7}=\pm 3(\text{3h}+\text{2k}+\text{6})\\ \mathrm{Taking}(+)\mathrm{ve}\mathrm{sign}:\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 9h}+\text{6k}-\text{7}=\text{9h}+\text{6k}+\text{18}\\ -7=18\\ \mathrm{Which}\text{is impossible.}\\ \mathrm{Taking}(-)\mathrm{ve}\mathrm{sign}:\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 9h}+\text{6k}-\text{7}=-\text{9h}-\text{6k}-\text{18}\\ \Rightarrow 18\mathrm{h}+12\mathrm{h}+11=0\\ \mathrm{Therefore},\text{the locus of equidistant point from given two}\\ \text{parallel lines is}18\mathrm{x}+12\mathrm{y}+11=0.\\ \mathrm{Thus},\text{the required equation of line is}18\mathrm{x}+12\mathrm{y}+11=0.\end{array}$

Q.73 A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Ans

BA is incident ray and AC is reflected ray on a point A on X-axis.

$\begin{array}{l}\mathrm{Let}\text{coordinates of point A on x-axis be}(\mathrm{a},0).\\ \angle \mathrm{BAP}=\angle \mathrm{CAP}\left[\mathrm{By}\text{the law of reflection}\right]\\ \Rightarrow \angle \mathrm{CAX}‘=\angle \mathrm{BAX}=\mathrm{\theta }\\ \therefore \angle \mathrm{CAX}=180\mathrm{°}-\mathrm{\theta }\\ \mathrm{Slope}\text{of BA}=\mathrm{tan\theta }\\ \frac{2-0}{1-\mathrm{a}}=\mathrm{tan\theta }\\ \Rightarrow \frac{2}{1-\mathrm{a}}=\mathrm{tan\theta }...\left(\mathrm{i}\right)\\ \mathrm{Slope}\text{of CA}=\tan (180\mathrm{°}-\mathrm{\theta })\\ \frac{3-0}{5-\mathrm{a}}=-\mathrm{tan\theta }\\ \frac{3}{5-\mathrm{a}}=-\mathrm{tan\theta }...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \frac{3}{5-\mathrm{a}}=-\frac{2}{1-\mathrm{a}}\\ \Rightarrow 3(1-\mathrm{a})=-2(5-\mathrm{a})\\ \Rightarrow 3-3\mathrm{a}=-10+2\mathrm{a}\\ \Rightarrow 5\mathrm{a}=-13\\ \Rightarrow \mathrm{a}=-\frac{13}{5}\\ \mathrm{So},\text{the coordinates of point A on x-axis is}(-\frac{13}{5},0).\end{array}$

Q.74

$\begin{array}{l}\mathbf{\text{Prove that the product of the lengths of the perpendiculars drawn from the points}}\left(\sqrt{{\text{a}}^{\text{2}}–{\text{b}}^{\text{2}}}\text{,0}\right)\\ \mathbf{\text{and}}(–\sqrt{{\text{a}}^{\text{2}}–{\text{b}}^{\text{2}}}\text{,0})\mathbf{\text{tothe line}}\frac{\mathbf{\text{x}}}{\mathbf{\text{a}}}\mathbf{\text{cosθ}}\mathbf{+}\frac{\mathbf{\text{y}}}{\mathbf{\text{b}}}\mathbf{\text{sinθ}}\mathbf{=}{\mathbf{\text{1 is b}}}^{\mathbf{\text{2}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\text{equation of line is:}\\ \frac{\mathrm{x}}{\mathrm{a}}\mathrm{cos\theta }+\frac{\mathrm{y}}{\mathrm{b}}\mathrm{sin\theta }=1...\left(\mathrm{i}\right)\\ \text{Distance of line}\left(\mathrm{i}\right)\text{from the point}(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2},0)\text{is}\\ {\text{d}}_{\text{1}}=\frac{|\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}\mathrm{cos\theta }+\frac{0}{\mathrm{b}}\text{sin}\mathrm{\theta }-1|}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\text{sin}}^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\\ =\frac{|\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}\mathrm{cos\theta }-1|}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\text{sin}}^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\\ {\text{d}}_{\text{1}}=\frac{\mathrm{b}|\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\mathrm{cos\theta }-\mathrm{a}|}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta }}}\\ \text{Distance of line}\left(\mathrm{i}\right)\text{from the point}(-\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2},0)\text{is}\end{array}$ $\begin{array}{l}{\text{d}}_{\text{2}}=\frac{|-\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}\mathrm{cos\theta }+\frac{0}{\mathrm{b}}\text{sin}\mathrm{\theta }-1|}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\text{sin}}^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\\ =\frac{|-\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}\mathrm{cos\theta }-1|}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\text{sin}}^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\\ =\frac{\mathrm{b}|\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\mathrm{cos\theta }+\mathrm{a}|}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta }}}\\ \mathrm{Now},\\ {\mathrm{d}}_1{\mathrm{d}}_2=\frac{\mathrm{b}|\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\mathrm{cos\theta }-\mathrm{a}|}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta }}}\times \frac{\mathrm{b}|\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\mathrm{cos\theta }+\mathrm{a}|}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta }}}\\ =\frac{{\mathrm{b}}^2\left|(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\mathrm{cos\theta }-\mathrm{a})(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\mathrm{cos\theta }+\mathrm{a})\right|}{({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta })}\\ =\frac{{\mathrm{b}}^2\left|\{({\mathrm{a}}^2-{\mathrm{b}}^2){\cos }^2\mathrm{\theta }-{\mathrm{a}}^2\}\right|}{({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta })}\\ =\frac{{\mathrm{b}}^2|{\mathrm{a}}^2{\cos }^2\mathrm{\theta }-{\mathrm{b}}^2{\cos }^2\mathrm{\theta }-{\mathrm{a}}^2|}{({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta })}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{{\mathrm{b}}^2|-{\mathrm{a}}^2(1-{\cos }^2\mathrm{\theta })-{\mathrm{b}}^2{\cos }^2\mathrm{\theta }|}{({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta })}\\ =\frac{{\mathrm{b}}^2|-{\mathrm{a}}^2{\sin }^2\mathrm{\theta }-{\mathrm{b}}^2{\cos }^2\mathrm{\theta }|}{({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta })}\\ {\mathrm{d}}_1{\mathrm{d}}_2=\frac{{\mathrm{b}}^2({\mathrm{a}}^2{\sin }^2\mathrm{\theta }+{\mathrm{b}}^2{\cos }^2\mathrm{\theta })}{({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\text{a}}^{\text{2}}{\text{sin}}^2\mathrm{\theta })}\\ ={\mathrm{b}}^2\\ \mathrm{Therefore},\text{the product of perpendicular distance of given two}\\ {\text{lines from two given points is b}}^{\text{2}}\text{.}\end{array}$

Q.75 A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

Ans

The equations of given lines are:
2x – 3y + 4 = 0 … (i)
3x + 4y – 5 = 0 … (ii)
6x – 7y + 8 = 0 … (iii)
Intersection point of line (i) and line (ii) is:

$\begin{array}{l}\Rightarrow (-\frac{1}{17},\frac{22}{17})\\ \mathrm{Slope}\text{of line (iii)}=-\frac{6}{-7}[\mathrm{m}=-\frac{\mathrm{A}}{\mathrm{B}}]\\ =\frac{6}{7}\\ \mathrm{Slope}\text{of perpendicular on line}\left(\mathrm{iii}\right)=-\frac{1}{\left(\frac{6}{7}\right)}=-\frac{7}{6}\end{array}$

$\begin{array}{l}\mathrm{Equation}\text{of line passing through}(-\frac{1}{17},\frac{22}{17})\text{and having}\\ \text{slope}-\frac{7}{6}\mathrm{is}:\\ \mathrm{y}-\frac{22}{17}=-\frac{7}{6}(\mathrm{x}+\frac{1}{17})\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow 17\mathrm{y}-22=-\frac{7}{6}(17\mathrm{x}+1)\\ \Rightarrow 102\mathrm{y}-132=-119\mathrm{x}-7\\ \Rightarrow 119\mathrm{x}+102\mathrm{y}-132+7=0\\ \Rightarrow 119\mathrm{x}+102\mathrm{y}-125=0\\ \mathrm{Thus},\text{the equation of the shortest path to reach from the}\\ \text{intersection point of line(i) and}\left(\mathrm{ii}\right)\text{to line}\left(\mathrm{iii}\right)\text{is}\\ 119\mathrm{x}+102\mathrm{y}-125=0.\end{array}$