CBSE Class 11 Maths Revision Notes Chapter 9

Q.1 Find the equations of the straight lines which passes through the origin and trisect the portion of the straight line [(x/a) + (y/b) = 1],
which is intercepted between the axes.

Ans

$\begin{array}{l}\text{The given line}\left[\right(\mathrm{x}/\mathrm{a})+(\mathrm{y}/\mathrm{b})=1]\text{makes an intercepts}\mathrm{a}\text{and}\mathrm{b}\text{on the axes.}\\ \text{Let the line cut the axes at}\mathrm{A}(\mathrm{a},0)\text{and}\mathrm{B}(0,\mathrm{b})\text{.}\\ \text{Let}\mathrm{P},\mathrm{Q}\text{be the points of trisection of}\mathrm{AB}\text{.}\\ \text{P divides}\mathrm{AB}\text{in the ratio}1:2\text{internally.}\\ \mathrm{P}=\left(\frac{1\left(0\right)+2\left(\mathrm{a}\right)}{1+2},\frac{1\left(\mathrm{b}\right)+2\left(0\right)}{1+2}\right)=\left(\frac{2\mathrm{a}}{3},\frac{\mathrm{b}}{3}\right)\\ \text{Q divides AB in the ratio}2:1\text{internally.}\\ \mathrm{Q}=\left(\frac{2\left(0\right)+1\left(\mathrm{a}\right)}{1+2},\frac{2\left(\mathrm{b}\right)+1\left(0\right)}{1+2}\right)=\left(\frac{\mathrm{a}}{3},\frac{2\mathrm{b}}{3}\right)\\ \text{The equation of the line joining origin and}\mathrm{P}\left(\frac{2\mathrm{a}}{3},\frac{\mathrm{b}}{3}\right)\text{is:}\\ \mathrm{y}-0=\frac{\frac{\mathrm{b}}{3}-0}{\frac{2\mathrm{a}}{3}-0}(\mathrm{x}-0)\\ \mathrm{y}=\frac{\mathrm{b}}{2\mathrm{a}}\mathrm{x}\Rightarrow \mathrm{bx}-2\mathrm{ay}=0\\ \text{The equation of the line joining origin and}\mathrm{Q}\left(\frac{\mathrm{a}}{3},\frac{2\mathrm{b}}{3}\right)\text{is:}\\ \mathrm{y}-0=\frac{\frac{2\mathrm{b}}{3}-0}{\frac{\mathrm{a}}{3}-0}(\mathrm{x}-0)\\ \mathrm{y}=\frac{2\mathrm{b}}{\mathrm{a}}\mathrm{x}\Rightarrow 2\mathrm{bx}-\mathrm{ay}=0\end{array}$

Q.2 Find the ratio in which the line joining (2, 3) and (4, 1) divides the line joining (1, 2) and (4, 3).

Ans

Let the given points be A (1, 2), B (4, 3), C (2, 3), D (4, 1).
Let CD divide AB in the ratio k : 1 internally at point P.

$\begin{array}{l}\mathrm{P}=\left(\frac{4\mathrm{k}+1}{\mathrm{k}+1},\frac{3\mathrm{k}+2}{\mathrm{k}+1}\right)\\ \text{The equation of}\mathrm{CD}\text{is :}\\ \mathrm{y}-3=\frac{(1-3)}{(4-2)}(\mathrm{x}-2)\\ \mathrm{y}-3=-(\mathrm{x}-2)\\ \mathrm{x}+\mathrm{y}=5\\ \text{Now,}\mathrm{P}\text{lies on CD.}\\ \frac{4\mathrm{k}+1}{\mathrm{k}+1}+\frac{3\mathrm{k}+2}{\mathrm{k}+1}=5\\ 4\mathrm{k}+1+3\mathrm{k}+2-5\mathrm{k}-5=0\\ 2\mathrm{k}-2=0\\ \text{The required ratio is}1:1.\end{array}$

Q.3 Find the equation of the angle bisectors of the angle between the coordinate axes.

Ans

Let l₁ and l₂ be the lines bisecting the angles between coordinate axes.
For l₁, m = tan 45° = 1 and c = 0.
The equation of l₁ is y = 1.x + 0 i.e. x – y = 0.
For l₂, m = tan 135° = –1 and c = 0.
The equation of l₂ is y = –1.x + 0 i.e. x + y = 0.

Q.4 Write the formula for the distance between two points.

Ans

Distance between two points P (x₁, y₁) and Q(x₂, y₂) is given by :
PQ = √(x₂ – x₁)² + (y_{2 –} y₁)²

Q.5 A line passes through (x₁,y₁) and (h,k). If the slope of the line is m, show that k – y₁ = m(h – x₁)

Ans

$\begin{array}{l}\text{If the slope of the line is}\mathrm{m}\text{, the equation of the line is :}\mathrm{y}=\mathrm{mx}+\mathrm{c}\\ \text{Since it passes through}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{, it will satisfy the equation of the line.}\\ {\mathrm{y}}_1={\mathrm{mx}}_1+\mathrm{c}...\left(\mathrm{i}\right)\\ \text{Since it passes through}(\mathrm{h},\mathrm{k})\text{, it will satisfy the equation of the line.}\\ \mathrm{k}=\mathrm{mh}+\mathrm{c}...\left(\mathrm{ii}\right)\\ \text{Subtracting (i) from (ii), we get}\\ \mathrm{k}-{\mathrm{y}}_1=\mathrm{mh}+\mathrm{c}-\left({\mathrm{mx}}_1+\mathrm{c}\right)\\ \mathrm{k}-{\mathrm{y}}_1=\mathrm{mh}+\mathrm{c}-{\mathrm{mx}}_1+\mathrm{c}\\ \mathrm{k}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{h}-{\mathrm{x}}_1\right)\\ \text{Hence proved.}\end{array}$

Q.6 Find the equation of the line joining the points (–3, –1) and (2, 3).
Also find the equation of the other line which is perpendicular to this line and passing through the point (5, 2).

Ans

$\begin{array}{l}\text{Let the given points be}\mathrm{P}(-3,-1),\mathrm{Q}(2,3)\text{and}\mathrm{R}(5,2).\\ \text{Equation of}\mathrm{PQ}\text{:}\\ \text{Slope of}\mathrm{PQ}=\frac{3-(-1)}{2-(-3)}=\frac{4}{5}\\ \text{Equation of}\mathrm{PQ}\text{is}\mathrm{y}-{\mathrm{y}}_1=\mathrm{Slope}\left(\mathrm{x}-{\mathrm{x}}_1\right)\\ \mathrm{y}-2=\frac{4}{5}[\mathrm{x}-(5\left)\right]\\ 5(\mathrm{y}-2)=4(\mathrm{x}-5)\\ 5\mathrm{y}-10=4\mathrm{x}-20\\ 4\mathrm{x}-5\mathrm{y}=10\\ \text{Let the slope of line}\mathrm{PQ}=\mathrm{m}\\ \text{Therefore,}\mathrm{m}\times \frac{4}{5}=-1\\ \Rightarrow \mathrm{m}=-\frac{5}{4}\\ (\mathrm{y}-2)=\frac{-5}{4}(\mathrm{x}-5)\\ 4\mathrm{y}-8=-5\mathrm{x}+25\\ 4\mathrm{y}+5\mathrm{x}=33\end{array}$

Q.7 Find a point on the x-axis which is equidistant from the points (5, 6) and (7, 8).

Ans

$\begin{array}{l}\text{Any point on the}\mathrm{X}\text{-axis would be of the form}(\mathrm{x},0).\\ \text{By distance formula, the distance between}(5,6)\text{and}(\mathrm{x},0)\text{would be}\\ \sqrt{{(5-\mathrm{x})}^2+{(6-0)}^2}...\left(\mathrm{i}\right)\\ \text{By distance formula, the distance between}(7,8)\text{and}(\mathrm{x},0)\text{would be}\\ \sqrt{{(7-\mathrm{x})}^2+{(8-0)}^2}...\left(\mathrm{ii}\right)\\ \text{As the points are equidistant, (i) and (ii) are equal.}\\ \sqrt{{(5-\mathrm{x})}^2+{(6-0)}^2}=\sqrt{{(7-\mathrm{x})}^2+{(8-0)}^2}\\ {\left(5-\mathrm{x}\right)}^2+{\left(6-0\right)}^2={\left(7-\mathrm{x}\right)}^2+{\left(8-0\right)}^2\\ 25+{\mathrm{x}}^2-10\mathrm{x}+36=49+{\mathrm{x}}^2-14\mathrm{x}+64\\ 14\mathrm{x}-10\mathrm{x}=113-61\\ 4\mathrm{x}=52\\ \mathrm{x}=13\\ \text{point is}(13,0)\end{array}$

Q.8

$\mathbf{\text{If the angle between two lines is}}{\mathbf{\pi }}_{\mathbf{/}\mathbf{4}}\mathbf{\text{and the slope of one of the lines is}}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\text{, find the slope of the other line.}}$

Ans

$\begin{array}{l}\text{The acute angle}\mathrm{\theta }\text{between two lines with slopes\hspace{0.33em}}{\mathrm{m}}_1\text{and}{\mathrm{m}}_2\text{is given by :}\\ \mathrm{tan\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ \text{Here,}{\mathrm{m}}_1=\frac{1}{3}\text{and}{\mathrm{m}}_2=\mathrm{m}\text{and}\mathrm{\theta }=\frac{\mathrm{\pi }}{4}.\\ \text{Putting these values in the above formula, we get}\\ \tan \frac{\mathrm{\pi }}{4}=\left|\frac{\mathrm{m}-\frac{1}{3}}{1+\frac{1}{3}\times \mathrm{m}}\right|=1\\ \mathrm{m}-\frac{1}{3}=1+\frac{\mathrm{m}}{3}\\ \mathrm{m}-\frac{\mathrm{m}}{3}=1+\frac{1}{3}\\ \frac{2\mathrm{m}}{3}=\frac{4}{3}\\ \mathrm{m}=2\end{array}$

Q.9 Find the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (3, 4).

Ans

The intercept form of a line is given by : [(x/a) + (y/b)] = 1,
where a and b are intercepts.
As there are equal intercepts, thus a = b.
Put this in the above equation.
[(x/a) + (y/a)] = 1
⇒ x + y = a
(3, 4) will satisfy the equation of the line.
Put (3, 4) in above equation we get
3 + 4 = a = 7
Thus, x + y = 7.

Q.10

$\begin{array}{l}\mathbf{\text{If}}\mathbf{p}\mathbf{\text{is the measure of the perpendicular segment from the origin on the line whose intercepts on the}}\\ \mathbf{\text{axes are}}\mathbf{a}\mathbf{\text{and}}\mathbf{b}\mathbf{\text{, show that\hspace{0.33em}}}\frac{\mathbf{1}}{{\mathbf{p}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{.}\end{array}$

Ans

$\begin{array}{l}\text{Let the given line cut the axes at}\mathrm{A}\text{and}\mathrm{B}\text{respectively.}\\ \mathrm{OA}=\mathrm{a}\text{and}\mathrm{OB}=\mathrm{b}.\\ \text{Let}\angle \mathrm{AOK}=\mathrm{\varphi }\text{and}\angle \mathrm{KOB}=90-\mathrm{\varphi }\\ \mathrm{In}\mathrm{\Delta OAK}\frac{\mathrm{OK}}{\mathrm{OA}}=\mathrm{cos\varphi }\mathrm{i}.\mathrm{e}.,\frac{\mathrm{p}}{\mathrm{a}}=\mathrm{cos\varphi }...\left(\mathrm{i}\right)\\ \text{In}\mathrm{\Delta OBK}\frac{\mathrm{OK}}{\mathrm{OB}}=\cos \left(90-\mathrm{\varphi }\right)=\mathrm{sin\varphi }\text{i.e.,}\\ \frac{\mathrm{p}}{\mathrm{b}}=\mathrm{sin\varphi }...\left(\mathrm{ii}\right)\\ \text{Squaring and adding (i) and (ii), we get}\\ {\left(\frac{\mathrm{p}}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{p}}{\mathrm{b}}\right)}^2={\left(\mathrm{cos\varphi }\right)}^2+{\left(\mathrm{sin\varphi }\right)}^2={\cos }^2\mathrm{\varphi }+{\sin }^2\mathrm{\varphi }=1\\ \Rightarrow \frac{1}{{\mathrm{p}}^2}=\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\end{array}$

Q.11 Find the distance between the parallel lines 2x – 5y + 7 = 0 and 2x – 5y + 5 = 0.

Ans

$\begin{array}{l}\text{Here, the equation of the line is of the form}\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0.\\ \text{Given that}\mathrm{A}=2,\mathrm{B}=-5,{\mathrm{C}}_1=7\text{and}{\mathrm{C}}_2=5\\ \mathrm{d}=\frac{\left|{\mathrm{C}}_1-{\mathrm{C}}_2\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\\ \mathrm{d}=\frac{|7-5|}{\sqrt{2^2+{(-5)}^2}}=\frac{\left|2\right|}{\sqrt{4+25}}=\frac{2}{\sqrt{29}}\end{array}$

Q.12 Find the distance of the point (5,-2) from the line 7x – 2y + 3 = 0.

Ans

$\begin{array}{l}\text{The perpendicular distance (d) of a line}\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0\text{from a point}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{is given by:}\\ \mathrm{d}=\frac{\left|{\mathrm{Ax}}_1+{\mathrm{By}}_1+\mathrm{C}\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\\ \mathrm{d}=\frac{\left|7\right(5)+-2(-2)+3|}{\sqrt{7^2+{(-2)}^2}}=\frac{|35+4+3|}{\sqrt{49+4}}=\frac{42}{\sqrt{53}}\end{array}$

Q.13

$\begin{array}{l}\text{Prove that the product of the lengths of the perpendiculars drawn from the points}\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2},0\right)\\ \text{and}\left(-\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2},0\right)\text{to the line}\frac{\mathrm{x}}{\mathrm{a}}\mathrm{cos\theta }+\frac{\mathrm{y}}{\mathrm{b}}\mathrm{sin\theta }=1\text{is}{\mathrm{b}}^2.\end{array}$

Ans

$\begin{array}{l}\text{The perpendicular distance (d) of a line}\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0\text{from a pointis}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{given by:}\\ \mathrm{d}=\frac{\left|{\mathrm{Ax}}_1+{\mathrm{By}}_1+\mathrm{C}\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\\ \text{Distance from}\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2},0\right)\text{to the given line is:}\\ {\mathrm{d}}_1=\frac{\frac{\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)}{\mathrm{a}}+\frac{\mathrm{sin\theta }}{\mathrm{b}}\left(0\right)-1}{\sqrt{{\left(\frac{\mathrm{cos\theta }}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{sin\theta }}{\mathrm{b}}\right)}^2}}=\frac{\mid \frac{\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}}{\mathrm{a}}}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}}}}\\ =\frac{\left|\frac{\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}}{\mathrm{a}}\right|}{\sqrt{\frac{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{a}}^2{\mathrm{b}}^2}}}=\frac{\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}}{\mathrm{a}}\times \frac{\mathrm{ab}}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}\\ =\frac{\mathrm{b}\left(\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}\right)}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}...\left(\mathrm{i}\right)\\ \text{Distance from}\left(-\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2},0\right)\text{to the given line is :}\\ {\mathrm{d}}_2=\frac{\frac{\mathrm{cos\theta }\left(-\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)}{\mathrm{a}}+\frac{\mathrm{sin\theta }}{\mathrm{b}}\left(0\right)-1}{\sqrt{{\left(\frac{\mathrm{Cos\theta }}{\mathrm{a}}\right)}^2+{\left(\frac{\mathrm{sin\theta }}{\mathrm{b}}\right)}^2}}=\frac{\frac{\mathrm{cos\theta }\left(-\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}}{\mathrm{a}}}{\sqrt{\frac{{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\\ =\frac{\frac{\mathrm{cos\theta }\left(-\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}}{\mathrm{a}}}{\sqrt{\frac{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{a}}^2{\mathrm{b}}^2}}}=\frac{-\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}}{\mathrm{a}}\times \frac{\mathrm{ab}}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}\\ =\frac{-\mathrm{b}\left(\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2}-{\mathrm{b}}^2\right)+\mathrm{a}\right)}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}\left(\mathrm{ii}\right)\\ \text{Their product}={\mathrm{d}}_1{\mathrm{d}}_2\\ =\frac{\mathrm{b}\left(\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)-\mathrm{a}\right)}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}\times \frac{-\mathrm{b}\left(\mathrm{cos\theta }\left(\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}\right)+\mathrm{a}\right)}{\sqrt{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}}\\ =\frac{-{\mathrm{b}}^2\left({\cos }^2\mathrm{\theta }\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)-{\mathrm{a}}^2\right)}{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}=\frac{-{\mathrm{b}}^2\left[{\mathrm{a}}^2\left({\cos }^2\mathrm{\theta }-1\right)-{\mathrm{b}}^2{\cos }^2\mathrm{\theta }\right]}{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}\\ =\frac{{\mathrm{b}}^2\left[{\mathrm{a}}^2\left(1-{\cos }^2\mathrm{\theta }\right)+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }\right]}{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}=\frac{{\mathrm{b}}^2\left({\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }\right)}{{\mathrm{b}}^2{\cos }^2\mathrm{\theta }+{\mathrm{a}}^2{\sin }^2\mathrm{\theta }}={\mathrm{b}}^2\end{array}$

Q.14 By using the concept of the equation of a line, prove that the three points (3,0), (-2,-2) and (8,2) are collinear.

Ans

$\begin{array}{l}\text{The equation of a line passing through the points}(3,0)\text{and}(-2,-2)\text{is}\\ \mathrm{y}-{\mathrm{y}}_1=\left(\mathrm{x}-{\mathrm{x}}_1\right)\times \frac{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}\\ \mathrm{y}-0=(\mathrm{x}-3)\times \frac{(-2-0)}{(-2-3)}\\ \mathrm{y}=(\mathrm{x}-3)\times \frac{(-2)}{(-5)}\\ 5\mathrm{y}=2\mathrm{x}-6\\ \text{Thus, the slope of the line is}\frac{2}{5}...\left(1\right)\\ \text{The equation of a line passing through the points}(8,2)\text{and}(-2,-2)\text{is}\\ \mathrm{y}-{\mathrm{y}}_1=\left(\mathrm{x}-{\mathrm{x}}_1\right)\times \frac{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}\\ \mathrm{y}-2=(\mathrm{x}-8)\times \frac{(-2-2)}{(-2-8)}\\ \mathrm{y}-2=(\mathrm{x}-8)\times \frac{(-4)}{(-10)}\\ 5\mathrm{y}-10=2\mathrm{x}-16\\ 2\mathrm{x}-5\mathrm{y}=6\\ \text{Thus, the slope of the line is}\frac{2}{5}...\left(2\right)\\ \text{From (1) and (2) the slopes of the lines are equal. Hence the given points are collinear.}\end{array}$

Q.15 Show that the path of a moving point such that its distances from two lines x – y = 5 and x + y = 5 are equal is a straight line.

Ans

$\begin{array}{l}\text{Let}(\mathrm{h},\mathrm{k})\text{is any point, whose distances from the two lines are equal.}\\ \text{The perpendicular distance (d) of a line}\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0\text{from a point}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{is given by ;}\\ \mathrm{d}=\frac{\left|{\mathrm{Ax}}_1+{\mathrm{By}}_1+\mathrm{C}\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\\ \mathrm{d}=\frac{|\mathrm{h}-\mathrm{k}-5|}{\sqrt{1+1}}=\frac{|\mathrm{h}+\mathrm{k}-5|}{\sqrt{1+1}}\\ \text{|}\mathrm{h}-\mathrm{k}-5|=|\mathrm{h}+\mathrm{k}-5|\\ \mathrm{h}-\mathrm{k}-5=\mathrm{h}+\mathrm{k}-5\\ 2\mathrm{k}=0\\ \mathrm{k}=0...\left(\mathrm{i}\right)\\ \text{and}\mathrm{h}-\mathrm{k}-5=-(\mathrm{h}+\mathrm{k}-5)\\ \mathrm{h}-\mathrm{k}-5=-\mathrm{h}-\mathrm{k}+5\\ 2\mathrm{h}=10\\ \mathrm{h}=5...\left(\mathrm{ii}\right)\\ \text{Thus, the point}(\mathrm{h},\mathrm{k})\text{satisfies the equations}\mathrm{y}=0\text{and}\mathrm{x}=5\text{, which represent straight lines.}\\ \text{Hence, the path of the point equidistant from the given lines is a straight line.}\end{array}$

Q.16 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

The intersection point of two lines 2x – y – 3 = 0 and 3x + y – 2 = 0
can be obtained by solving these equations :
3x + y – 2 = 0
2x – y – 3 = 0
———————–
5x = 5
x = 1 and y = 2 – 3x = 2 – 3 = –1.
Since the point (1, –1) lies on the line px + 2y – 3 = 0,
it will satisfy the equation of the line.
p(1) + 2(–1) – 3 = 0
p – 2 – 3 = 0
p = 5

Q.17 Find the equation of the straight line which makes an angle of 60° with the positive direction of x-axis and cuts an intercept of 6 units on the y – axis.

Ans

We have, m = tan 60° = (1/√3) and c = + 6.
The equation of the line is given by :
y = mx + c Putting values, y = (1/√3)x + 6
(√3)y = x + 6√3
x – (√3)y + 6√3 = 0

Q.18 Find the equation of the straight line which passes through the point (–2, 3) and making an angle of 30° with the positive direction of x- axis. Also find the points on the line which are at the distance of
(i) 3 units
(ii) 5 units from the point (–2, 3).

Ans

$\begin{array}{l}\text{The given point}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{is}(-2,3)\text{and}\mathrm{\theta }=30°.\\ \text{The equation of the line is given by :}\frac{\mathrm{x}-{\mathrm{x}}_1}{\mathrm{cos\theta }}=\frac{\mathrm{y}-{\mathrm{y}}_1}{\mathrm{sin\theta }}=\mathrm{r}\\ \text{where}\mathrm{r}\text{is the distance between}(\mathrm{x},\mathrm{y})\text{and}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{. This is symmetric form.}\\ \frac{\mathrm{x}-(-2)}{\cos 30°}=\frac{\mathrm{y}-3}{\sin 30°}=\mathrm{r}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}……………}\left(1\right)\\ \sin 30°(\mathrm{x}+2)=\cos 30°(\mathrm{y}-3)\\ \frac{1}{2}(\mathrm{x}+2)=\frac{\sqrt{3}}{2}(\mathrm{y}-3)\\ (\mathrm{x}+2)=\sqrt{3}(\mathrm{y}-3)\\ \mathrm{x}+2=\sqrt{3}\mathrm{y}-3\sqrt{3}\\ \mathrm{x}-\sqrt{3}\mathrm{y}+(2+3\sqrt{3})=0\\ \text{This is the equation of the line.}\\ \left(\text{i}\right)\text{\hspace{0.33em}The points on the line which are at the distance of}3\text{units from\hspace{0.33em}the point}(-2,3)\\ \text{are}\mathrm{r}=3\text{\hspace{0.33em}From (1),}\\ \frac{\mathrm{x}-(-2)}{\cos 30°}=3\text{and}\frac{\mathrm{y}-3}{\sin 30°}=3\\ \mathrm{x}=\frac{3\sqrt{3}}{2}-2\Rightarrow \mathrm{x}=\frac{3\sqrt{3}-4}{2}\\ \mathrm{y}=\frac{5}{2}+3\Rightarrow \mathrm{y}=\frac{11}{2}\\ \text{Thus the point}\left(\frac{5\sqrt{3}-4}{2},\frac{11}{2}\right)\text{is the required point.}\end{array}$

Q.19 What is the slope of a vertical line and a horizontal line?

Ans

The slope (m) of a horizontal line is 0 and that of a vertical line is undefined.

Q.20 What is the relation between the slopes of two lines when they are
(i) Parallel (ii) Perpendicular

Ans

Let m₁ and m₂ be the slopes of the two given lines.
(i) When they are parallel, then m₁ = m₂.
(ii) When they are perpendicular, then m₁m₂ = –1.

Q.21 Which of the lines 2x + 7y – 9 = 0 and 4x – y + 11 = 0 is farther from the point (3,4)?

Ans

$\begin{array}{l}\text{The perpendicular distance (d) of a line}\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0\\ \text{from a point}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{is given by :}\\ \mathrm{d}=\frac{\left|{\mathrm{Ax}}_1+{\mathrm{By}}_1+\mathrm{C}\right|}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\\ \text{Distance of the point from the line}2\mathrm{x}+7\mathrm{y}-9=0\text{is}\\ {\mathrm{d}}_1=\frac{2\left(3\right)+7\left(4\right)-9}{\sqrt{2^2+7^2}}=\frac{\mid 6+28-9}{\sqrt{49+4}}=\frac{25}{\sqrt{53}}=3.43\\ \text{Distance of the point from the line}2\mathrm{x}+7\mathrm{y}-9=0\text{is}\\ {\mathrm{d}}_2=\frac{\left|4\right(3)+(-1\left)\right(4)+11|}{\sqrt{4^2+{(-1)}^2}}=\frac{|12-4+11|}{\sqrt{16+1}}=\frac{19}{\sqrt{17}}=4.608\\ \text{Hence, the line}4\mathrm{x}-\mathrm{y}+11=0\text{is farther from the point}(3,4)\end{array}$

Q.22 If the lines 7x + y – 3 = 0, 5x + ky – 5 = 0 and 8x – y – 2 = 0 are concurrent, then find the value of k.

Ans

Three lines are said to be concurrent if they intersect at one point.
Thus, we find the intersection point of two lines and say that it lies on the third line and must satisfy the equation of the third line.
On solving 7x + y – 3 = 0 and 8x – y – 2 = 0, we get x = (1/3) and y = (2/3).
Now, (1/3,2/3) lies on 5x + ky – 5 = 0. Thus,
(5/3) + (2/3)k – 5 = 0
(2/3)k = 5 – (5/3) = (10/3)
k = 5

Q.23 Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line 2x + 3y – 5 = 0.

Ans

$\begin{array}{l}\text{Let}\mathrm{Q}(\mathrm{x},\mathrm{y})\text{be the image of the point}\mathrm{P}(1,2)\text{in the line}2\mathrm{x}+3\mathrm{y}-5=0.\\ \text{Thus, the given line is the perpendicular bisector of the line segment}\mathrm{PQ}.\\ \text{Slope of}\mathrm{PQ}\times \text{Slope of line}2\mathrm{x}+3\mathrm{y}-5=0=-1\\ \frac{\mathrm{y}-2}{\mathrm{x}-1}\times \frac{-2}{3}=-1\\ 2\mathrm{y}-4=3\mathrm{x}-3\\ 3\mathrm{x}-2\mathrm{y}=-1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}………}\left(1\right)\\ \text{Also the midpoint of PQ i.e.}\left(\frac{\mathrm{x}+1}{2},\frac{\mathrm{y}+2}{2}\right)\text{will satisfy the\hspace{0.33em}given\hspace{0.33em}\hspace{0.33em}equation}.\\ \text{So,}\\ 2\left(\frac{\mathrm{x}+1}{2}\right)+3\left(\frac{\mathrm{y}+2}{2}\right)-5=0\\ 2\mathrm{x}+2+3\mathrm{y}+6-10=0\\ 2\mathrm{x}+3\mathrm{y}=2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}…………}..\left(2\right)\\ \text{Solving (1) and (2), we get}\mathrm{x}=\frac{1}{13},\mathrm{y}=\frac{8}{13}\text{.}\\ \text{Thus the point is}\left(\frac{1}{13},\frac{8}{13}\right).\end{array}$

Q.24 A line passes through (2, 4) and the sum of the intercepts on the axes is 12. Find the equation of the line.

Ans

$\begin{array}{l}\text{Let the}\mathrm{x}\text{-intercept of the line}\mathrm{be}=\mathrm{a}\text{and the}\mathrm{y}\text{-intercept will be}(12-\mathrm{a})\text{.}\\ \text{Thus the equation of the line in the intercept form is}\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{12-\mathrm{a}}=1\left[\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\right]\\ \text{This line passes through}(2,4).\\ \text{Thus, it will satisfy the above equation.}\\ \frac{2}{\mathrm{a}}+\frac{4}{12-\mathrm{a}}=1\\ 24-2\mathrm{a}+4\mathrm{a}=12\mathrm{a}-{\mathrm{a}}^2\\ {\mathrm{a}}^2-10\mathrm{a}+24=0\\ (\mathrm{a}-6)(\mathrm{a}-4)=0\\ \mathrm{a}=6,4\\ \text{Case}1:\text{When}\mathrm{a}=6,\text{the equation becomes :}\\ \frac{\mathrm{x}}{6}+\frac{\mathrm{y}}{12-6}=1\\ \mathrm{x}+\mathrm{y}=6\\ \text{Case}2\text{:\hspace{0.33em}When}\mathrm{a}=4,\text{the equation becomes :}\\ \frac{\mathrm{x}}{4}+\frac{\mathrm{y}}{12-4}=1\\ 2\mathrm{x}+\mathrm{y}=8\end{array}$

Q.25 A line forms a triangle in the first quadrant with the coordinate axes. If the area of the triangle is 54√3 sq. units and the perpendicular drawn from the origin to the line makes an angle of 60° with the X-axis, find the equation of the line.

Ans

Let ∠POA = 60°, OP = p, OA = a, OB = b.
In ΔOPA, p/a = cos 60° = 1/2 ⇒ a = 2p
In ΔOPB, p/b = cos 30° = √3/2 ⇒ b = 2p/ √3
Area of ΔAOB = (1/2)ab = (1/2) x 2p x (2p/√3) = (2/√3)p²
Thus,
(2/√3)p² = 54√3
p² = 81
p = 9

Using, x cos θ + y sin θ = p , we get
x cos 60° + y sin 60° = 9
x + √3y – 18 = 0

Q.26 Find the equation of the straight line which passes through the origin and making an angle of 60° with the line 3x + 4y = 2.

Ans

$\begin{array}{l}\text{Slope of the given line}3\mathrm{x}+4\mathrm{y}=2\text{is}(-3)/4.\text{Let the slope of the required line be}\mathrm{m}\text{.}\\ \mathrm{tan\theta }=\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|\\ \text{Here,}{\mathrm{m}}_1=\frac{-3}{4}\text{and}{\mathrm{m}}_2=\mathrm{m}\text{and}\mathrm{\theta }={60}^{°}.\text{Putting these values in the above\hspace{0.33em}formula,}\\ \tan 60°=\left|\frac{\mathrm{m}-\frac{-3}{4}}{1+\frac{-3}{4}\times \mathrm{m}}\right|=\sqrt{3}\\ \mathrm{m}+\frac{3}{4}=\sqrt{3}\left(1-\frac{3\mathrm{m}}{4}\right)\text{\hspace{0.33em}and}\mathrm{m}+\frac{3}{4}=-\sqrt{3}\left(1-\frac{3\mathrm{m}}{4}\right)\\ \mathrm{m}+\frac{3\sqrt{3}\mathrm{m}}{4}=\sqrt{3}-\frac{3}{4}\text{\hspace{0.33em}and}\mathrm{m}+\frac{3\sqrt{3}\mathrm{m}}{4}=-\sqrt{3}+\frac{3}{4}\\ \mathrm{m}\frac{(4+3\sqrt{3})}{4}=\frac{4\sqrt{3}-3}{4}\text{\hspace{0.33em}and}\mathrm{m}\frac{(4+3\sqrt{3})}{4}=\frac{-4\sqrt{3}+3}{4}\\ \mathrm{m}=\frac{4\sqrt{3}-3}{4+3\sqrt{3}}\text{\hspace{0.33em}and}\mathrm{m}=\frac{-4\sqrt{3}+3}{4+3\sqrt{3}}\\ \text{Using}\mathrm{y}=\mathrm{mx}+\mathrm{c}\text{and}\mathrm{c}=0\text{(as the line passes through the origin), we get}\\ \mathrm{y}=\left(\frac{4\sqrt{3}-3}{4+3\sqrt{3}}\right)\mathrm{x}+0=\left(\frac{4\sqrt{3}-3}{4+3\sqrt{3}}\right)\mathrm{x}\text{and}\mathrm{y}=\left(\frac{-4\sqrt{3}+3}{4+3\sqrt{3}}\right)\mathrm{x}+0=\left(\frac{-4\sqrt{3}+3}{4+3\sqrt{3}}\right)\mathrm{x}\\ (4\sqrt{3}-3)\mathrm{x}-(4+3\sqrt{3})\mathrm{y}=0\text{and}(-4\sqrt{3}+3)\mathrm{x}-(4+3\sqrt{3})\mathrm{y}=0\end{array}$

Q.27 Find the coordinates of the centroid of a triangle whose vertices are (1,6), (–1,2) and (3,1).

Ans

$\begin{array}{l}\text{We know that the coordinates of the centroid\hspace{0.33em}of a triangle whose vertices are}\\ \left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right),\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)\text{is}\\ \left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3}\right)\\ \text{So, the coordinates of the centroid of a\hspace{0.33em}triangle whose vertices are}(1,6),(-1,12)\\ \text{and}(3,1)\text{is}\\ \left(\frac{1-1+3}{3},\frac{6+2+1}{3}\right)=\left(\frac{3}{3},\frac{9}{3}\right)=(1,3)\end{array}$

Q.28 Find the distance between the points (acosA + bsinA,0) and Q(0, asinA – BcosA).

Ans

$\begin{array}{l}\text{The distance}\\ \mathrm{PQ}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ {\left.\mathrm{PQ}=\sqrt{{(\mathrm{acosA}+\mathrm{bsinA})}^2+(\mathrm{bcosA}-\mathrm{asinA}}\right)}^2\\ \mathrm{PQ}=\sqrt{{\mathrm{a}}^2\left({\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}\right)+{\mathrm{b}}^2\left({\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}\right)}\\ \mathrm{PQ}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\end{array}$

Q.29 Find the slope of the straight line which makes an angle of 135° with +ve direction of X-axis.

Ans

$\begin{array}{l}\text{Slope,}\mathrm{m}=\tan 135°\\ =\tan \left(180°-45°\right)\\ =-\tan 45°\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=}-\text{1}\end{array}$

Q.30 Find the slope of the straight line which passes through the point (2,–3) and (7, 7).

Ans

$\begin{array}{l}\text{Slope,}\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\\ =\frac{7-(-3)}{7-2}\\ =\frac{7+3}{5}\\ =\frac{10}{5}\\ =2\end{array}$

Q.31 The three vertices of a parallelogram taken in the order are (–1,0), (3,1) and (2,2) respectively. Find the coordinates of the forth vertex and hence, find its area.

Ans

$\begin{array}{l}\text{Let}\mathrm{A}(-1,0),\mathrm{B}(3,1),\mathrm{C}(2,2)\text{and}\mathrm{D}(\mathrm{x},\mathrm{y})\text{be the vertices of the parallelogram taken in order.}\\ \text{Since, the diagonals of a parallelogram bisect each other}\\ \therefore \text{\hspace{0.33em}\hspace{0.33em}coordinates of mid-point of}\mathrm{AC}=\text{coordinates of mid point of}\mathrm{BD}\\ \Rightarrow \left(\frac{-1+2}{2},\frac{0+2}{2}\right)=\left(\frac{3+\mathrm{x}}{2},\frac{1+\mathrm{y}}{2}\right)\\ \Rightarrow \frac{1}{2}=\frac{3+\mathrm{x}}{2}\\ \Rightarrow 2\mathrm{x}+6=2\\ \Rightarrow \mathrm{x}=\frac{-4}{2}\\ \Rightarrow \mathrm{x}=-2\\ \text{and\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1=\frac{1+\mathrm{y}}{2}\\ \Rightarrow \mathrm{y}+1=2\\ \Rightarrow \mathrm{y}=1\\ \text{Hence, the fourth vertex of parallelogram is}(-2,1)\\ \text{Area of}\mathrm{\Delta }=\frac{1}{2}\left|\left\{{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right\}\right|\\ \text{Area of}\mathrm{\Delta ABD}=\frac{1}{2}\left|\right\{-1(1-1)+3(1-0)-2(0-1)\left\}\right|\\ \text{Area of}\mathrm{\Delta ABD}=\frac{1}{2}(0+3+2)=\frac{5}{2}...\left(1\right)\\ \because \mathrm{\Delta ABD}\cong \mathrm{\Delta BCD}\\ \therefore \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Area of}\mathrm{\Delta ABD}=\text{Area of}\mathrm{\Delta BCD}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Area of}\mathrm{\Delta BCD}=\frac{5}{2}...\left(2\right)\\ \text{Using\hspace{0.33em}equation\hspace{0.33em}}\left(\text{1}\right)\text{\hspace{0.33em}and\hspace{0.33em}}\left(\text{2}\right)\text{,}\\ \text{Area of}\mathrm{\Delta ABCD}=\text{Area of}\mathrm{\Delta ABD}+\text{Area of}\mathrm{\Delta BCD}\\ =\frac{5}{2}+\frac{5}{2}\\ =\frac{10}{2}\\ =5\text{square\hspace{0.33em}units.}\end{array}$

Q.32

$\begin{array}{l}\mathbf{\text{Show that the equation of the line passing\hspace{0.33em}through the origin and making an angle}}\mathbf{\theta }\\ \mathbf{\text{with the line}}\mathbf{y}\mathbf{=}\mathbf{mx}\mathbf{+}\mathbf{c}\mathbf{\text{is}}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{=}\mathbf{\pm }\frac{\mathbf{m}\mathbf{+}\mathbf{tan\theta }}{\mathbf{1}\mathbf{-}\mathbf{mtan\theta }}\end{array}$

Ans

$\begin{array}{l}\text{Let the slope of the required line be}{\mathrm{m}}^{‘}\\ \text{Since this line makes an angle}\mathrm{\theta }\text{with line}\\ \mathrm{y}=\mathrm{mx}+\mathrm{c}\\ \Rightarrow \mathrm{tan\theta }=\left|\frac{{\mathrm{m}}^{‘}-\mathrm{m}}{1+{\mathrm{m}}^{‘}\mathrm{m}}\right|\\ \Rightarrow \mathrm{tan\theta }=\pm \frac{{\mathrm{m}}^{‘}-\mathrm{m}}{1+{\mathrm{m}}^{‘}\mathrm{m}}\\ \text{on taking\hspace{0.33em}+\hspace{0.33em}ve sign}\\ \Rightarrow \mathrm{tan\theta }+{\mathrm{m}}^{‘}\mathrm{m}\mathrm{tan\theta }={\mathrm{m}}^{‘}-\mathrm{m}\\ \Rightarrow \mathrm{m}+\mathrm{tan\theta }={\mathrm{m}}^{‘}-{\mathrm{m}}^{‘}\mathrm{m}\mathrm{tan\theta }\\ \Rightarrow {\mathrm{m}}^{‘}(1-\mathrm{m}\mathrm{tan\theta })=\mathrm{m}+\mathrm{tan\theta }\\ \Rightarrow {\mathrm{m}}^{‘}=\frac{\mathrm{m}+\mathrm{tan\theta }}{1-\mathrm{m}\mathrm{tan\theta }}\\ \text{on taking -ve sign}\\ \Rightarrow \mathrm{tan\theta }+{\mathrm{m}}^{‘}\mathrm{m}\mathrm{tan\theta }=-{\mathrm{m}}^{‘}+\mathrm{m}\\ \Rightarrow {\mathrm{m}}^{‘}(1+\mathrm{m}\mathrm{tan\theta })=\mathrm{m}-\mathrm{tan\theta }\\ \Rightarrow {\mathrm{m}}^{‘}=\frac{\mathrm{m}-\mathrm{tan\theta }}{1+\mathrm{m}\mathrm{tan\theta }}\\ \Rightarrow {\mathrm{m}}^{‘}=\pm \frac{\mathrm{m}+\mathrm{tan\theta }}{1-\mathrm{m}\mathrm{tan\theta }}\\ \text{The equation of the line passing through}(0,0)\\ \text{and having slope},{\mathrm{m}}^{‘}=\pm \frac{\mathrm{m}+\mathrm{tan\theta }}{1-\mathrm{m}\mathrm{tan\theta }}\text{is}\\ \mathrm{y}-0={\mathrm{m}}^{‘}(\mathrm{x}-0)\\ \Rightarrow \mathrm{y}=\pm \frac{\mathrm{m}+\mathrm{tan\theta }}{1-\mathrm{m}\mathrm{tan\theta }}\mathrm{x}\\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\pm \frac{\mathrm{m}+\mathrm{tan\theta }}{1-\mathrm{m}\mathrm{tan\theta }}\end{array}$