NCERT Solutions Class 10 Maths Chapter 7

Q.1 Find the distance between following pairs of points:
(i) (2, 3), (4, 1)
(ii) (–5, 7), (–1, 3)
(iii) (a, b), (–a, –b)

Ans.

$\begin{array}{l}\text{(i)}\\ {\text{Distance between the two points (x}}_1,{\text{y}}_1){\text{and (x}}_2,{\text{y}}_2\text{) is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}.\\ \text{Therefore, distance between points (}2,3)\text{and (}4,1)\\ =\sqrt{{\left(4-2\right)}^2+{\left(1-3\right)}^2}=\sqrt{4+4}=2\sqrt{2}\text{units}\\ \text{(ii)}\\ {\text{Distance between the two points (x}}_1,{\text{y}}_1){\text{and (x}}_2,{\text{y}}_2)\text{is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}.\\ \text{Therefore, distance between points (}-5,7)\text{and (}-1,3)\\ =\sqrt{{\left(-5-(-1)\right)}^2+{\left(7-3\right)}^2}=\sqrt{16+16}=4\sqrt{2}\text{units}\\ \text{(iii)}\\ {\text{Distance between the two points (x}}_1,{\text{y}}_1){\text{and (x}}_2,{\text{y}}_2)\text{is}\\ \text{given by the expression}\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}.\\ \text{Therefore, distance between points (a},\mathrm{b})\text{and (}-\mathrm{a},-\mathrm{b})\\ =\sqrt{{\left(\mathrm{a}-(-\mathrm{a})\right)}^2+{\left(\mathrm{b}-(-\mathrm{b})\right)}^2}=\sqrt{4{\mathrm{a}}^2+4{\mathrm{b}}^2}=2\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\end{array}$

Q.2 Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Ans.

$\begin{array}{l}\text{Distance between points (}0,0)\text{and (}36,15)\\ =\sqrt{{(0-36)}^2+{(0-15)}^2}=\sqrt{1296+225}=\sqrt{1521}=39\text{\hspace{0.17em}\hspace{0.17em}units}\\ \text{Yes, we can find the distance between the given towns A and B.}\\ \text{Let town A be at (0, 0) and town B be at (36, 15). Then, as per}\\ \text{above calculation, distance between the two towns is}39\text{\hspace{0.17em}\hspace{0.17em}km}.\end{array}$

Q.3 Determine if the points (1, 5), (2, 3) and (– 2, –11) are collinear.

Ans.

$\begin{array}{l}\text{Let the given points are P(1, 5), Q(2, 3) and R(}-2,-11).\\ \text{Now, using distance formula we find distance between these}\\ \text{points i.e., PQ, QR and PR.}\\ \text{Distance between points P(1, 5) and Q(2, 3)}\\ =\text{PQ}=\sqrt{{(1-2)}^2+{(5-3)}^2}=\sqrt{1+4}=\sqrt{5}\\ \text{Distance between points Q(2, 3) and R(}-2,-11)\\ =\text{QR}=\sqrt{{(2+2)}^2+{(3+11)}^2}=\sqrt{16+196}=\sqrt{212}=2\sqrt{53}\\ \text{Distance between points P(1, 5) and R(}-2,-11)\\ =\text{PR}=\sqrt{{(1+2)}^2+{(5+11)}^2}=\sqrt{9+256}=\sqrt{265}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{QR}=\sqrt{5}+2\sqrt{53}\ne \sqrt{265}=\text{PR}\\ \text{i.e., PQ}+\text{QR}\ne \text{PR}\\ \text{Therefore, points P(1, 5), Q(2, 3) and R(}-2,-11)\text{are}\\ \text{not collinear.}\end{array}$

Q.4 Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Ans.

$\begin{array}{l}\text{Let the given points are P(5,}-2\text{), Q(6, 4) and R(7},-2).\\ \text{Now, using distance formula we find distance between these}\\ \text{points i.e., PQ, QR and PR.}\\ \text{Distance between points P(5,}-2\text{) and Q(6, 4)}\\ =\text{PQ}=\sqrt{{(5-6)}^2+{(-2-4)}^2}=\sqrt{1+36}=\sqrt{37}\\ \text{Distance between points Q(6, 4) and R(7},-2)\\ =\text{QR}=\sqrt{{(6-7)}^2+{(4+2)}^2}=\sqrt{1+36}=\sqrt{37}\\ \text{Distance between points P(5,}-2\text{) and R(7},-2)\\ =\text{PR}=\sqrt{{(5-7)}^2+{(-2+2)}^2}=\sqrt{4+0}=2\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{QR}=\sqrt{37}+\sqrt{37}\ne 2=\text{PR}\\ \text{i.e., PQ}+\text{QR}\ne \text{PR}\\ \text{Therefore, points P(5,}-2\text{), Q(6, 4) and R(7},-2)\text{are}\\ \text{not collinear. They form an isosceles triangle as lengths of}\\ \text{sides PQ and QR are equal.}\end{array}$

Q.5 In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.


Ans.

$\begin{array}{l}\text{It is given that 4 friends are seated at the points A(3, 4),}\\ \text{B(6, 7), C(9, 4) and D(6, 1).}\\ \text{Now, using distance formula we find length of each side}\\ \text{of the quadrilatral ABCD and diagonals AC and BD.}\\ \text{AB}=\sqrt{{\left(6-3\right)}^2+{\left(7-4\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{BC}=\sqrt{{\left(6-9\right)}^2+{\left(7-4\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{CD}=\sqrt{{\left(9-6\right)}^2+{\left(4-1\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{DA}=\sqrt{{\left(6-3\right)}^2+{\left(4-1\right)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\\ \text{AC}=\sqrt{{\left(9-3\right)}^2+{\left(4-4\right)}^2}=\sqrt{36+0}=6,\\ \text{BD}=\sqrt{{\left(6-6\right)}^2+{\left(7-1\right)}^2}=\sqrt{0+36}=6.\\ \text{We see that sides AB, BC, CD and DA are equal in lengths.}\\ \text{Also, diagonals AC and BD are equal.}\\ \text{Therefore, quadrilateral ABCD is a square. Hence, Champa}\\ \text{is correct.}\end{array}$

Q.6 Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Let the given points are A}(-1,-2),\text{B}(1,0),\text{C}(-1,2)\text{and}\\ \text{D}(-3,0).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{(-1-1)}^2+{(-2-0)}^2}=\sqrt{4+4}=2\sqrt{2},\\ \text{BC}=\sqrt{{(1+1)}^2+{(0-2)}^2}=\sqrt{4+4}=2\sqrt{2},\\ \text{CD}=\sqrt{{(-1+3)}^2+{(2-0)}^2}=\sqrt{4+4}=2\sqrt{2},\\ \text{DA}=\sqrt{{(-1+3)}^2+{(-2-0)}^2}=\sqrt{4+4}=2\sqrt{2}\\ \text{and}\\ \text{AC}=\sqrt{{(-1+1)}^2+{(-2-2)}^2}=\sqrt{0+16}=4,\\ \text{BD}=\sqrt{{(1+3)}^2+{(0-0)}^2}=\sqrt{16+0}=4.\\ \text{We see that sides AB, BC, CD and DA are equal in lengths.}\\ \text{Also, diagonals AC and BD are equal.}\\ \text{Therefore, the given points form a square.}\\ \text{(ii)}\\ \text{Let the given points are A}(-3,\text{5}),\text{B}(3,\text{1}),\text{C}(0,\text{3})\text{and}\\ \text{D}(-1,-4).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{(-3-3)}^2+{(5-1)}^2}=\sqrt{36+16}=\sqrt{52},\\ \text{BC}=\sqrt{{(3-0)}^2+{(1-3)}^2}=\sqrt{9+4}=\sqrt{13},\\ \text{CD}=\sqrt{{(0+1)}^2+{(3+4)}^2}=\sqrt{1+49}=5\sqrt{2},\\ \text{DA}=\sqrt{{(-1+3)}^2+{(-4-5)}^2}=\sqrt{4+81}=\sqrt{85}\\ \text{We see that sides AB, BC, CD and DA are not equal.}\\ \text{Therefore, the given points do not form a special}\\ \text{quadrilateral. These points form a general quadrilateral.}\\ \text{(iii)}\\ \text{Let the given points are A}(4,5),\text{B}(7,6),\text{C}(4,3)\text{and D}(1,2).\\ \text{Now, using distance formula we have}\\ \text{AB}=\sqrt{{(4-7)}^2+{(5-6)}^2}=\sqrt{9+1}=\sqrt{10},\\ \text{BC}=\sqrt{{(7-4)}^2+{(6-3)}^2}=\sqrt{9+9}=3\sqrt{2},\\ \text{CD}=\sqrt{{(4-1)}^2+{(3-2)}^2}=\sqrt{9+1}=\sqrt{10},\\ \text{DA}=\sqrt{{(4-1)}^2+{(5-2)}^2}=\sqrt{9+9}=3\sqrt{2}\\ \text{and}\\ \text{AC}=\sqrt{{(4-4)}^2+{(5-3)}^2}=\sqrt{0+4}=2,\\ \text{BD}=\sqrt{{(7-1)}^2+{(6-2)}^2}=\sqrt{36+16}=\sqrt{52}.\\ \text{We see that opposite sides of quadrilateral ABCD are equal.}\\ \text{But, diagonals AC and BD are not equal.}\\ \text{Therefore, the given points form a parallelogram.}\end{array}$

Q.7 Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Ans.

$\begin{array}{l}\text{Let T(x, 0) be the point on the x-axis which is equidistant from}\\ \text{A(2,}-\text{5) and B(}-\text{2, 9)}.\\ \text{Now, using distance formula we have}\\ \text{TA}=\sqrt{{(\mathrm{x}-2)}^2+{(0+5)}^2}=\sqrt{{\mathrm{x}}^2-4\mathrm{x}+4+25}=\sqrt{{\mathrm{x}}^2-4\mathrm{x}+29}\\ \text{and}\\ \text{TB}=\sqrt{{(\mathrm{x}+2)}^2+{(0-9)}^2}=\sqrt{{\mathrm{x}}^2+4\mathrm{x}+4+81}=\sqrt{{\mathrm{x}}^2+4\mathrm{x}+85}\\ \text{Now,}\\ \text{TA}=\text{TB}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{{\mathrm{x}}^2-4\mathrm{x}+29}=\sqrt{{\mathrm{x}}^2+4\mathrm{x}+85}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^2-4\mathrm{x}+29={\mathrm{x}}^2+4\mathrm{x}+85\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\mathrm{x}=29-85=-56\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-7\\ \text{Hence, the required point is (}-7,\text{0).}\end{array}$

Q.8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Ans.

$\begin{array}{l}\text{It is given that the distance between the points P(2,}-3)\text{and}\\ \text{(10, y) is 10 units.}\\ \text{Therefore, using distance formula we have}\\ \sqrt{{(2-10)}^2+{(-3-\mathrm{y})}^2}=10\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}64+9+6\mathrm{y}+{\mathrm{y}}^2=100\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}^2+6\mathrm{y}-27=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}^2+9\mathrm{y}-3\mathrm{y}-27=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}(\mathrm{y}+9)-3(\mathrm{y}+9)=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(\mathrm{y}-3)(\mathrm{y}+9)=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=3\text{or \hspace{0.17em}}\mathrm{y}=-9\end{array}$

Q.9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Ans.

$\begin{array}{l}\text{It is given that Q(0, 1) is equidistant from P(5,}-3)\text{and R(x, 6).}\\ \text{Therefore, using distance formula, we have}\\ \sqrt{{(0-5)}^2+{(1+3)}^2}=\sqrt{{(0-\mathrm{x})}^2+{(1-6)}^2}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25+16={\mathrm{x}}^2+25\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^2=16\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\pm 4\\ \text{When}\mathrm{x}=4,\text{then}\\ \text{QR}=\sqrt{{(0-4)}^2+{(1-6)}^2}=\sqrt{41}\text{units}\\ \text{and}\\ \text{PR}=\sqrt{{(5-4)}^2+{(-3-6)}^2}=\sqrt{82}\text{units}\\ \text{When}\mathrm{x}=-4,\text{\hspace{0.17em}\hspace{0.17em}then}\\ \text{QR}=\sqrt{{(0+4)}^2+{(1-6)}^2}=\sqrt{41}\text{units}\\ \text{and}\\ \text{PR}=\sqrt{{(5+4)}^2+{(-3-6)}^2}=9\sqrt{2}\text{units}\end{array}$

Q.10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

Ans.

$\begin{array}{l}\text{It is given that the point (x, y) is equidistant from the}\\ \text{point (3,}6)\text{and R(}-3\text{, 4).}\\ \text{Therefore, using distance formula, we have}\\ \sqrt{{(\mathrm{x}-3)}^2+{(\mathrm{y}-6)}^2}=\sqrt{{(\mathrm{x}+3)}^2+{(\mathrm{y}-4)}^2}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-12\mathrm{y}+36={\mathrm{x}}^2+6\mathrm{x}+9+{\mathrm{y}}^2-8\mathrm{y}+16\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-6\mathrm{x}-6\mathrm{x}-12\mathrm{y}+8\mathrm{y}+45-25=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-12\mathrm{x}-4\mathrm{y}+20=0\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+\mathrm{y}-5=0\end{array}$

Q.11

$\text{Find the coordinates of the point which divides the join of (}-\text{1, 7) and (4,}-\text{3) in the ratio 2}:\text{3.}$

Ans.

$\begin{array}{l}\text{We find the coordinates of the point which divides the}\\ \text{join of (}-\text{1, 7) and (4,}-\text{3) in the ratio 2}:\text{3 using section}\\ \text{formula.}\\ \text{So, we have}\\ \mathrm{x}=\frac{2\times 4+3(-1)}{2+3}=1\\ \text{and}\\ \mathrm{y}=\frac{2\times (-3)+3\times 7}{2+3}=3\\ \text{Therefore, coordinates of the required point are (1, 3).}\end{array}$

Q.12 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Ans.

$\begin{array}{l}{\text{Let P(x}}_1{\text{, y}}_1{\text{) and Q(x}}_2{\text{, y}}_2\text{) are the points of trisection of}\\ \text{the line segment joining the given points A(4,}-1)\text{and}\\ \text{B(}-2,-3).\\ \text{So, AP}=\text{PQ}=\text{BQ and}\\ \text{Therefore, point P divides AB internally in the ratio 1}:2.\\ \text{By section formula, we have}\\ {\mathrm{x}}_1=\frac{1\times (-2)+2\times 4}{1+2}=2\\ {\mathrm{y}}_1=\frac{1\times (-3)+2\times (-1)}{1+2}=\frac{-5}{3}\\ \text{Therefore, coordinates of the point P are}(2,\frac{-5}{3}).\\ \text{Point Q divides AB internally in the ratio 2}:1.\text{\hspace{0.17em}So, we have}\\ {\mathrm{x}}_2=\frac{2\times (-2)+1\times 4}{2+1}=0\\ \text{and}\\ {\mathrm{y}}_2=\frac{2\times (-3)+1\times (-1)}{2+1}=\frac{-7}{3}\\ \text{Therefore, coordinates of the point Q are}\left(\text{0,}\frac{-7}{3}\right)\text{.}\end{array}$

Q.13

\begin{array}{l}\text{To conduct Sports Day activities, in your rectangular}\\ \text{shaped school ground ABCD, lines have been drawn}\\ \text{with chalk powder at a distance of 1m each}\text{. 100 flower}\\ \text{pots have been placed at a distance of 1m from each}\\ \text{other along AD, as shown in the following figure}\text{. Niharika}\\ \text{runs}\frac{1}{4}\text{th the distance AD on the 2nd line and posts a}\\ \text{green flag}\text{. Preet runs}\frac{\text{1}}{5}\text{th the distance AD on the eighth}\\ \text{line and posts a red flag}\text{. What is the distance between}\\ \text{both the flags? If Rashmi has to post a blue flag exactly}\\ \text{halfway between the line segment joining the two flags,}\\ \text{where should she post her flag?}\end{array}

Ans.

$\begin{array}{l}\text{Niharika posts the green flag at}\frac{1}{4}\text{of the distance AD i.e.,}\frac{100}{4}\text{m}=25\text{m}\\ \text{from the starting point of 2nd line.}\\ \text{Therefore, the coordinates of this point G are (2, 25).}\\ \text{Also, Preet posts red flag at}\frac{1}{5}\text{of the distance AD i.e.,}\frac{100}{5}\text{m}=20\text{m}\\ \text{from the starting point of 8th line.}\\ \text{Therefore, the coordinates of this point R are (8, 20).}\\ \text{Distance between these flags}=\sqrt{{(2-8)}^2+{(25-20)}^2}\\ =\sqrt{36+25}=\sqrt{61}\text{m}\\ \text{Rashmi has to post blue flag at the mid-point A(x, y) of the line segment}\\ \text{joining the red and green flags.}\\ \text{So, we have}\\ \mathrm{x}=\frac{2+8}{2}=5\text{and}\mathrm{y}=\frac{25+20}{2}=\frac{45}{2}=22.5\\ \text{Coordinates of the point where blue flag should be posted are (5, 22.5).}\\ \text{Therefore, Rashmi should post her blue flag at 22.5 m on the 5th line.}\end{array}$

Q.14 Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).

Ans.

$\begin{array}{l}\text{Let the ratio in which the line segment joining the points}\\ (-3,10)\text{and}(6,-8)\text{is divided by}(-1,6)\text{be}\mathrm{k}:1.\\ \text{Therefore,}\\ -1=\frac{6\mathrm{k}-3}{\mathrm{k}+1}\\ \text{or}-\mathrm{k}-1=6\mathrm{k}-3\\ \text{or}-\mathrm{k}-6\mathrm{k}-1+3=0\\ \text{or}-7\mathrm{k}+2=0\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{2}{7}\\ \text{Therefore, the required ratio is 2:7.}\end{array}$

Q.15 Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Ans.

$\begin{array}{l}\text{Let the ratio in which the line segment joining A}(1,-5)\\ \text{and B}(-4,5)\text{is divided by the x-axis be}\mathrm{k}:1.\\ \text{Therefore,}\\ \mathrm{x}=\frac{-4\mathrm{k}+1}{\mathrm{k}+1}\text{and}0=\frac{5\mathrm{k}-5}{\mathrm{k}+1}\\ \text{Now,}\\ 0=\frac{5\mathrm{k}-5}{\mathrm{k}+1}\\ \text{or}0=5\mathrm{k}-5\\ \text{or \hspace{0.17em}}\mathrm{k}=1\\ \text{Therefore, the required ratio is 1:1.}\\ \text{Also,}\\ \mathrm{x}=\frac{-4\mathrm{k}+1}{\mathrm{k}+1}=\frac{-4+1}{1+1}=\frac{-3}{2}\\ \text{Therefore, the given line segment is divided by the point}\\ (\frac{-3}{2},\text{0})\text{in the ratio 1:1.}\end{array}$

Q.16 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Ans.

$\begin{array}{l}\text{Let A}\left(\text{1, 2}\right)\text{, B}\left(\text{4, y}\right)\text{, C}\left(\text{x, 6}\right)\text{and D}\left(\text{3, 5}\right)\text{are the vertices}\\ \text{of a parallelogram ABCD. Diagonals of a parallelogram}\\ \text{bisect each other. Therefore, intersection point O of}\\ \text{diagonals AC and BD is the mid-point of these diagonals.}\\ \text{Now, O is the mid-point of AC.}\\ \text{Therefore, coordinates of O are}(\frac{1+\mathrm{x}}{2},\frac{2+6}{2})\text{i.e.,}(\frac{1+\mathrm{x}}{2},4).\\ \\ \text{Also, O is the mid-point of BD.}\\ \text{Therefore, coordinates of O are}(\frac{4+3}{2},\frac{\mathrm{y}+5}{2})\text{i.e.,}(\frac{7}{2},\frac{\mathrm{y}+5}{2}).\\ \text{So, we have}\\ \frac{1+\mathrm{x}}{2}=\frac{7}{2}\text{i.e.,}\mathrm{x}=6\\ \text{and}\\ \frac{\mathrm{y}+5}{2}=4\text{i.e.,}\mathrm{y}=3\end{array}$

Q.17 Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).

Ans.

$\begin{array}{l}\text{Let the coordinates of point A be (x, y).}\\ \text{Mid-point of AB is (2,}-\text{3), which is the centre of the circle.}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} (2,}-\text{3)}=(\frac{\mathrm{x}+1}{2},\frac{\mathrm{y}+4}{2})\\ \text{i.e., 2}=\frac{\mathrm{x}+1}{2}\text{and}-\text{3}=\frac{\mathrm{y}+4}{2}\\ \text{i.e.,}\mathrm{x}=3\text{and}\mathrm{y}=-10\\ \text{Therefore, the coordinates of A are (3,}-10).\end{array}$

Q.18

$\begin{array}{l}\text{If A and B are (}-\text{2,}-\text{2) and (2,}-\text{4), respectively,}\\ \text{find the coordinates of P such that AP}=\frac{3}{7}\text{AB and P}\\ \text{lies on the line segment AB.}\end{array}$

Ans.

$\begin{array}{l}\text{Point P divides the line segment AB in the ratio}3:4.\\ \text{Therefore,}\\ \text{coordinates of P}=(\frac{3\times 2+4\times (-2)}{3+4},\frac{3\times (-4)+4\times (-2)}{3+4})\\ =(\frac{-2}{7},\frac{-20}{7})\end{array}$

Q.19

$\begin{array}{l}\text{Find the coordinates of the points which divide the}\\ \text{line segment joining A(}-\text{2, 2) and B(2, 8) into four}\\ \text{equal parts.}\end{array}$

Ans.

$\begin{array}{l}\text{From the above figure, we observe that points P, Q, R divide}\\ \text{the line segment in ratio}1:3,\text{1:1, 3:1 respectively.}\\ \text{Therefore,}\\ \text{coordinates of P}=(\frac{2\times 1-2\times 3}{1+3},\frac{1\times 8+3\times 2}{1+3})=(-1,\frac{7}{2}),\\ \text{coordinates of Q}=(\frac{2-2}{1+1},\frac{8+2}{1+1})=(0,5),\\ \text{and}\\ \text{coordinates of R}=(\frac{3\times 2-2\times 1}{1+3},\frac{3\times 8+1\times 2}{1+3})=(1,\frac{13}{2}).\end{array}$

Q.20

$\begin{array}{l}\text{Find the area of a rhombus if its vertices are (3, 0),}\\ \text{(4, 5), (}-\text{1, 4) and (}-\text{2,}-\text{1) taken in order.}\\ \text{[Hint : Area of a rhombus}=\frac{1}{2}\text{(product of its diagonals)]}\end{array}$

Ans.

$\begin{array}{l}\text{Let A}\left(\text{3, 0}\right)\text{, B}\left(\text{4, 5}\right)\text{, C}(-1\text{, 4})\text{and D}(-\text{2,}-1)\text{are the vertices}\\ \text{of a rhombus ABCD.}\end{array}$

$\begin{array}{l}\text{Length of diagonal AC}=\sqrt{{\left[3-(-1)\right]}^2+{(0-4)}^2}=4\sqrt{2}\\ \text{Length of diagonal BD}=\sqrt{{\left[4-(-2)\right]}^2+{[5-(-1\left)\right]}^2}=6\sqrt{2}\\ \\ \text{Therefore, area of rhombus ABCD}=\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}\\ =24\text{square units.}\end{array}$

Q.21

$\begin{array}{l}\text{Find the area of the triangle whose vertices are :}\\ \text{(i) (}2,3),(-1,0),(2,-4)\\ \text{(ii)}(-5,-1),(3,-5),(5,2)\end{array}$

Ans.

$\begin{array}{l}\text{Area of a triangle with vertives (}{\mathrm{x}}_1{\text{, y}}_1\text{), (}{\mathrm{x}}_2\text{,}{\mathrm{y}}_2\text{) and (}{\mathrm{x}}_3\text{,}{\mathrm{y}}_3\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right\}\text{.}\\ \text{Therefore,}\\ \text{(i) Area of the triangle with vertices (}2,3),(-1,0\left)\text{and}\right(2,-4)\\ =\frac{1}{2}\left\{2(0-(-4\left)\right)+(-1)(-4-3)+2(3-0)\right\}\\ =\frac{1}{2}(8+7+6)=\frac{21}{2}\text{square units}\\ \text{(ii) Area of the triangle with vertices}(-5,-1),(3,-5)\text{and}(5,2)\\ =\frac{1}{2}\left\{-5(-5-2)+3(2-(-1\left)\right)+5(-1-(-5)\right\}\\ =\frac{1}{2}(35+9+20)=32\text{square units}\end{array}$

Q.22 In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, –5)

Ans.

$\begin{array}{l}\text{Points (}{\mathrm{x}}_1{\text{, y}}_1\text{), (}{\mathrm{x}}_2\text{,}{\mathrm{y}}_2\text{) and (}{\mathrm{x}}_3\text{,}{\mathrm{y}}_3\text{) are collinear if}\\ \frac{1}{2}\left\{{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right\}=0\\ \text{Therefore,}\\ \text{(i) Points (7,}-\text{2)},\text{(5, 1) and (3, k) are collinear if}\\ \frac{1}{2}\left\{7(1-\mathrm{k})+5(\mathrm{k}+2)+3(-2-1)\right\}=0\\ \text{or 7}-\text{7}\mathrm{k}+5\mathrm{k}+10-9=0\\ \text{or}-\text{2k}+\text{8}=0\\ \text{or}\mathrm{k}=4\\ \text{(ii) Points (8, 1)},\text{(}\mathrm{k}\text{,}-4\text{) and (2,}-5\text{) are collinear if}\\ \frac{1}{2}\left[8\{-4-(-5\left)\right\}+\mathrm{k}(-5-1)+2\{1-(-4\left)\right\}\right]=0\\ \text{or 8}-6\mathrm{k}+10=0\\ \text{or}-\text{6k}=-1\text{8}\\ \text{or}\mathrm{k}=3\end{array}$

Q.23

$\begin{array}{l}\text{Find the area of the triangle formed by joining the}\\ \text{mid-points of the sides of the triangle whose}\\ \text{vertices are}(0,-1),(2,1)\text{and (0, 3). Find the}\\ \text{ratio of this area to the area of the given triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{Area of a triangle with vertices (}{\mathrm{x}}_1{\text{, y}}_1\text{), (}{\mathrm{x}}_2\text{,}{\mathrm{y}}_2\text{) and (}{\mathrm{x}}_3\text{,}{\mathrm{y}}_3\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right\}\text{.}\\ \text{Therefore,}\\ \text{area of the triangle with vertices (0},-1),(2,\text{1}\left)\text{and}\right(0,3)\\ =\frac{1}{2}\left\{0(1-3)+2(3+1)+0(-1-1)\right\}\\ =\frac{1}{2}(0+8+0)=4\text{square units}\\ \text{Vertices of the triangle formed by joining the mid-points of the}\\ \text{sides of the triangle with vertices}(0,-1),(2,1)\text{and (0, 3)}\\ \text{are}(\frac{0+2}{2},\frac{-1+1}{2}),(\frac{2+0}{2},\frac{1+3}{2})\text{and}(\frac{0+0}{2},\frac{-1+3}{2})\text{i.e.,}\\ (1,0),(1,2)\text{and}(0,1).\\ \text{Area of the triangle with vertices}(1,0),(1,2)\text{and}(0,1)\\ =\frac{1}{2}\left\{1(2-1)+1(1-0)+0(0-2)\right\}\\ =\frac{1}{2}(1+1+0)=1\text{square unit}\\ \text{Therefore, required ratio}=1:4\end{array}$

Q.24

$\begin{array}{l}\text{Find the area of the quadrilateral whose vertices, taken in order, are (}-\text{4,}-\text{2), (}-\text{3,}-\text{5), (}3,-2\text{)}\\ \text{and (2, 3).}\end{array}$

Ans.

\begin{array}{l}\text{Area of a triangle with vertices (}{x}_1{\text{, y}}_1\text{), (}{x}_2\text{,}{y}_2\text{) and (}{x}_3\text{,}{y}_3\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right\}\text{.}\\ \text{Therefore,}\\ \text{area of}\Delta \text{ABC with vertices A(}-4,-2),\text{B(}-3,-5)\text{and C(3,}-2)\\ =\frac{1}{2}\left\{-4(-5+2)+(-3)(-2+2)+3(-2+5)\right\}\\ =\frac{1}{2}(12+0+9)=\frac{21}{2}\text{square units}\\ \text{area of}\Delta \text{ADC with vertices A(}-4,-2),\text{D(2},\text{3})\text{and C(3,}-2)\\ =\frac{1}{2}\left\{-4(3+2)+2(-2+2)+3(-2-3)\right\}\\ =\frac{1}{2}(-20+0-15)=-\frac{35}{2}\\ \text{But area is always a positive quantity}\text{.}\\ \therefore \text{area of}\Delta \text{ADC}=\frac{35}{2}\text{square units}\\ \text{Area of the given quadrilateral ABCD}\\ =\text{area of}\Delta \text{ABC}+\text{area of}\Delta \text{ADC}\\ =\left(\frac{21}{2}+\frac{35}{2}\right)\text{square units}\\ =\text{28 square units}\end{array}

Q.25

$\begin{array}{l}\mathrm{You}\mathrm{have}\mathrm{studied}\mathrm{in}\mathrm{Class}\mathrm{IX},\left(\mathrm{Chapter}9,\mathrm{Example}3\right),\\ \mathrm{that}\mathrm{a}\mathrm{median}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{divides}\mathrm{it}\mathrm{into}\mathrm{two}\mathrm{triangles}\\ \mathrm{of}\mathrm{equal}\mathrm{areas}.\mathrm{Verify}\mathrm{this}\mathrm{result}\mathrm{for}\mathrm{\Delta ABC}\mathrm{whose}\\ \mathrm{vertices}\mathrm{are}\mathrm{A}\left(4,-6\right),\mathrm{B}\left(3,-2\right)\mathrm{and}\mathrm{C}\left(5,2\right).\end{array}$

Ans.

$\begin{array}{l}\text{Area of a triangle with vertives (}{\mathrm{x}}_1{\text{, y}}_1\text{), (}{\mathrm{x}}_2\text{,}{\mathrm{y}}_2\text{) and (}{\mathrm{x}}_3\text{,}{\mathrm{y}}_3\text{)}\\ \text{is given by the expression}\frac{1}{2}\left\{{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right\}\text{.}\\ \text{Therefore,}\\ \text{area of}\mathrm{\Delta }\text{ABC with vertices A(}4,-6),\text{B(}3,-2)\text{and C(5,}2)\\ =\frac{1}{2}\left\{4(-2-2)+3(2+6)+5(-6+2)\right\}\\ =\frac{1}{2}(-16+24-20)=-6\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta }\text{ABC}=6\text{square units}\\ \text{Now, D is the mid-point of BC. Therefore, coordinates of D are}\\ (\frac{3+5}{2},\frac{-2+2}{2})\text{i.e., (4, 0).}\\ \text{area of}\mathrm{\Delta }\text{ADC with vertices A(}4,-6),\text{D(4},\text{0})\text{and C(5,}2)\\ =\frac{1}{2}\left\{4(0-2)+4(2+6)+5(-6+0)\right\}\\ =\frac{1}{2}(-8+32-30)=-3\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta }\text{ADC}=3\text{square units}\\ \text{area of}\mathrm{\Delta }\text{ABD with vertices A(}4,-6),\text{B(3,}-2)\text{and D(4},\text{0})\\ =\frac{1}{2}\left\{4(-2-0)+3(0+6)+4(-6+2)\right\}\\ =\frac{1}{2}(-8+18-16)=-3\\ \text{But area is always a positive quantity.}\\ \therefore \text{area of}\mathrm{\Delta }\text{ABD}=3\text{square units}\\ \text{Now, we observe that}\\ \text{area of}\mathrm{\Delta }\text{ADC}=\text{area of}\mathrm{\Delta }\text{ABD}=3\text{square units}\\ \text{Hence, we verified the result}\mathbf{that}\mathbf{a}\mathbf{median}\mathbf{of}\mathbf{a}\mathbf{triangle}\mathbf{divides}\\ \mathbf{it}\mathbf{into}\mathbf{two}\mathbf{triangles}\mathbf{of}\mathbf{equal}\mathbf{areas}.\end{array}$

Q.26

$\begin{array}{l}\text{Determine the ratio in which the line\hspace{0.33em}}2\mathrm{x}+\mathrm{y}-4=0\text{\hspace{0.33em}divides the line segment joining the points A(}2,-2\text{)}\\ \text{and B(3, 7).\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\text{Let the given line divide the line segment joining the points}\\ \text{A(2,}-2)\text{and B(3, 7) in the ratio}\mathrm{k}:1.\\ \text{Coordinates of the point of division}=(\frac{3\mathrm{k}+2}{\mathrm{k}+1},\frac{7\mathrm{k}-2}{\mathrm{k}+1})\\ \text{This point of division lies on the line 2}\mathrm{x}+\mathrm{y}-4=0.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\frac{3\mathrm{k}+2}{\mathrm{k}+1}+\frac{7\mathrm{k}-2}{\mathrm{k}+1}-4=0\\ \text{or 6}\mathrm{k}+4+7\mathrm{k}-2-4\mathrm{k}-4=0\\ \text{or 9}\mathrm{k}-2=0\\ \text{or}\mathrm{k}=\frac{2}{9}\\ \text{Therefore, the required ratio is 2:9}\end{array}$

Q.27

$\text{Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.}$

Ans.

$\begin{array}{l}\text{Points (}{\mathrm{x}}_1{\text{, y}}_1\text{), (}{\mathrm{x}}_2\text{,}{\mathrm{y}}_2\text{) and (}{\mathrm{x}}_3\text{,}{\mathrm{y}}_3\text{) are collinear if}\\ {\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)=0\\ \text{Therefore,}\\ \text{Points (x,}\mathrm{y}\text{)},\text{(1, 2) and (7, 0) are collinear if}\\ \mathrm{x}(2-0)+1(0-\mathrm{y})+7(\mathrm{y}-2)=0\\ \text{or}2\mathrm{x}-\mathrm{y}+7\mathrm{y}-14=0\\ \text{or}2\mathrm{x}+6\mathrm{y}-14=0\\ \text{or}\mathrm{x}+3\mathrm{y}-7=0\end{array}$

Q.28 Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).

Ans.

$\begin{array}{l}\text{Let O(x, y) be the centRE of the circle and A(6,}-6),\text{B(3,}-7)\\ \text{and (3, 3) are points on the circle.}\\ \text{Then OA, OB and OC are radii of the circle and}\\ \text{OA}=\text{OB}=\text{OC.}\\ \text{Therefore, by distance formula, we have}\\ \sqrt{{(\mathrm{x}-6)}^2+{(\mathrm{y}+6)}^2}=\sqrt{{(\mathrm{x}-3)}^2+{(\mathrm{y}+7)}^2}\\ \text{or}{\mathrm{x}}^2-12\mathrm{x}+36+{\mathrm{y}}^2+12\mathrm{y}+36={\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2+14\mathrm{y}+49\\ \text{or}-6\mathrm{x}-2\mathrm{y}+14=0\\ \text{or}-3\mathrm{x}-\mathrm{y}+7=0\\ \text{or}\mathrm{y}=-3\mathrm{x}+7...\text{(1)}\\ \text{Also,}\\ \text{OA}=\text{OC}\\ \text{or}\sqrt{{(\mathrm{x}-6)}^2+{(\mathrm{y}+6)}^2}=\sqrt{{(\mathrm{x}-3)}^2+{(\mathrm{y}-3)}^2}\\ \text{or}{\mathrm{x}}^2-12\mathrm{x}+36+{\mathrm{y}}^2+12\mathrm{y}+36={\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-6\mathrm{y}+9\\ \text{or}-6\mathrm{x}+18\mathrm{y}+54=0\\ \text{or}-\mathrm{x}+3\mathrm{y}+9=0\\ \text{or}-\mathrm{x}+3(-3\mathrm{x}+7)+9=0\text{[From (1),}\mathrm{y}=-3\mathrm{x}+7\text{]}\\ \text{or}-\mathrm{x}-9\mathrm{x}+21+9=0\\ \text{or}-\text{10}\mathrm{x}+30=0\\ \text{or}-\mathrm{x}+3=0\\ \text{or}\mathrm{x}=3\\ \text{On putting this value of x in (1), we get}\\ \\ \mathrm{y}=-3\times \left(3\right)+7=-2\\ \text{Therefore, coordinates of the centre are (}3,-2).\end{array}$

Q.29 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.


Ans.

$\begin{array}{l}\text{Let ABCD be a square having A(}-1,2\left)\text{and C(3, 2}\right)\text{as opposite}\\ {\text{vertices. Let other two opposite vertices are B(x, y) and D(x}}_1,{\text{y}}_1).\\ \text{We know that sides of a square are equal to each other.}\\ \therefore \text{AB}=\text{BC}\\ \text{or}\sqrt{{(\mathrm{x}+1)}^2+{(\mathrm{y}-2)}^2}=\sqrt{{(\mathrm{x}-3)}^2+{(\mathrm{y}-2)}^2}\\ \text{or}{\mathrm{x}}^2+2\mathrm{x}+1+{(\mathrm{y}-2)}^2={\mathrm{x}}^2-6\mathrm{x}+9+{(\mathrm{y}-2)}^2\\ \text{or}8\mathrm{x}=8\\ \text{or}\mathrm{x}=1\\ \text{We know that all interior angles are of 90° in a square.}\\ \text{Therefore, using Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we have}\\ {\text{AB}}^2+{\text{BC}}^2={\text{AC}}^2\\ \text{or}{(\mathrm{x}+1)}^2+{(\mathrm{y}-2)}^2+{(\mathrm{x}-3)}^2+{(\mathrm{y}-2)}^2={(3+1)}^2+{(2-2)}^2\\ {\text{or x}}^2+2\mathrm{x}+1+{\mathrm{y}}^2-4\mathrm{y}+4+{\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-4\mathrm{y}+4=16\\ {\text{or 2x}}^2+2{\mathrm{y}}^2-4\mathrm{x}+2-8\mathrm{y}=0\\ \text{or 2}+2{\mathrm{y}}^2-4+2-8\mathrm{y}=0\text{[}\mathrm{x}=1]\\ {\text{or 2y}}^2-8\mathrm{y}=0\\ {\text{or y}}^2-4\mathrm{y}=0\\ \text{or y(y}-4)=0\\ \text{i.e., y}=0\text{or y}=4\\ \text{We know that in a square, the diagonals are of equal length}\\ \text{and bisect each other at 90°. Let O be the mid-point of AC.}\\ \text{Therefore, it is also the mid-point of BD.}\\ \text{Coordinates of point O}=(\frac{-1+3}{2},\frac{2+2}{2})=(\frac{\mathrm{x}+{\mathrm{x}}_1}{2},\frac{\mathrm{y}+{\mathrm{y}}_1}{2})\\ \text{i.e.,}(\frac{\mathrm{x}+{\mathrm{x}}_1}{2},\frac{\mathrm{y}+{\mathrm{y}}_1}{2})=(1,2)\\ \text{i.e.,}\frac{\mathrm{x}+{\mathrm{x}}_1}{2}=1\text{and}\frac{\mathrm{y}+{\mathrm{y}}_1}{2}=2\\ \text{i.e.,}\mathrm{x}+{\mathrm{x}}_1=2\text{and}\mathrm{y}+{\mathrm{y}}_1=4\\ \text{Putting the values of x and y, we get}{\mathrm{x}}_1=1\text{and}{\mathrm{y}}_1=0\text{or}{\mathrm{y}}_1=4.\\ \text{Therefore, the required coordinates of other two opposite vertices}\\ \text{of the given square are (1, 0) and (1, 4).}\end{array}$

Q.30 The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown
in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?


Ans.

$\begin{array}{l}\text{(i)}\\ \text{We take A as origin and AD as x-axis and AB as y-axis.}\\ \text{From the given figure, we observe that the coordinates}\\ \text{of point P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.}\\ \text{Now,}\\ \text{Area of}\mathrm{\Delta }\text{PQR}=\frac{1}{2}\left[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right]\\ =\frac{1}{2}\left[4(2-5)+3(5-6)+6(6-2)\right]\\ =\frac{1}{2}[-12-3+24]\\ =\frac{9}{2}\text{square units}\\ \text{(ii)}\\ \text{We take C as origin and CB as x-axis and CD as y-axis.}\\ \text{From the given figure, we observe that the coordinates}\\ \text{of point P, Q and R are (12, 2), (13, 6) and (10, 3) respectively.}\\ \text{Now,}\\ \text{Area of}\mathrm{\Delta }\text{PQR}=\frac{1}{2}\left[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right]\\ =\frac{1}{2}\left[12(6-3)+13(3-2)+10(2-6)\right]\\ =\frac{1}{2}[36+13-40]\\ =\frac{9}{2}\text{square units}\\ \text{We observe that area of the triangle in both cases is same.}\end{array}$

Q.31

$\begin{array}{l}\text{The vertices of a}\mathrm{\Delta }\text{\hspace{0.17em}ABC are A(4, 6), B(1, 5) and C(7, 2).}\\ \text{A line is drawn to intersect sides AB and AC at D and E}\\ \text{respectively, such that}\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{1}}{4}.\text{Calculate the area}\\ \text{of the}\mathrm{\Delta }\text{\hspace{0.17em}ADE and compare it with the area of}\mathrm{\Delta }\text{\hspace{0.17em}ABC.}\\ \text{(Recall converse of basic proportionality theorem and}\\ \text{Theorem 6.6 related to the ratio of the areas of two}\\ \text{similar triangles).}\end{array}$

Ans.

$\begin{array}{l}\text{Given that,}\\ \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\\ \text{or}\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{AE}+\mathrm{EC}}=\frac{1}{4}\\ \text{or}\frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{AE}}=4\\ \text{or}\frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}-1=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{AE}}-1=4-1\\ \text{or}\frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{AE}}=3\\ \text{or}\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{1}{3}\\ \text{Therefore, D and E are two points on side AB and AC respectively}\\ \text{such that they divide side AB and AC in the ratio 1:3.}\\ \text{Coordinates of point D}=(\frac{1\times 1+3\times 4}{1+3},\frac{1\times 5+3\times 6}{1+3})\\ =(\frac{13}{4},\frac{23}{4})\\ \text{Coordinates of point E}=(\frac{1\times 7+3\times 4}{1+3},\frac{1\times 2+3\times 6}{1+3})\\ =(\frac{19}{4},\frac{20}{4})\\ \text{Area of a triangle}=\frac{1}{2}\left[{\mathrm{x}}_1({\mathrm{y}}_2-{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3-{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right]\\ \text{Area of}\mathrm{\Delta }\text{ADE\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}[4(\frac{23}{4}-\frac{20}{4})+\frac{13}{4}(\frac{20}{4}-6)+\frac{19}{4}(6-\frac{23}{4})]\\ =\frac{1}{2}[3-\frac{13}{4}+\frac{19}{16}]\\ =\frac{1}{2}\left[\frac{48-52+19}{16}\right]\\ =\frac{15}{32}\text{square units}\\ \text{Area of}\mathrm{\Delta }\text{ABC\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]\\ =\frac{1}{2}[12-4+7]\\ =\frac{15}{2}\text{\hspace{0.17em}\hspace{0.17em}square units}\\ \text{Ratio between the areas of}\mathrm{\Delta }\text{ADE and}\mathrm{\Delta }\text{ABC is 1:16.}\end{array}$

Q.32

$\begin{array}{l}\text{Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of}\mathrm{\Delta }\text{ABC.}\\ \text{(i) The median from A meets BC at D. Find the}\\ \text{coordinates of the point D.}\\ \text{(ii) Find the coordinates of the point P on AD such that}\\ \text{AP}:\text{PD}=\text{2}:\text{1}\\ \text{(iii) Find the coordinates of points Q and R on medians}\\ \text{BE and CF respectively such that BQ}:\text{QE}=\text{2}:\text{1}\\ \text{and CR}:\text{RF}=\text{2}:\text{1.}\\ \text{(iv) What do yo observe?}\\ \text{[Note: The point which is common to all the three}\\ \text{medians is called the centroid and this point divides}\\ \text{each median in the ratio 2}:\text{1.]}\\ \text{(v) If A(}{\mathrm{x}}_1,{\mathrm{y}}_1\text{), B(}{\mathrm{x}}_2,{\mathrm{y}}_2\text{) and C(}{\mathrm{x}}_3,{\mathrm{y}}_3\text{) are the vertices of}\\ \mathrm{\Delta }\text{ABC, find the coordinates of the centroid of the}\\ \text{triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Median AD of}\mathrm{\Delta }\text{ABC bisects BC. So, D is the mid-point of}\\ \text{side BC.}\\ \therefore \text{Coordinates of D}=(\frac{6+1}{2},\frac{5+4}{2})=(\frac{7}{2},\frac{9}{2})\\ \text{(ii)}\\ \text{Point P divides the side AD in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of P}=(\frac{2\times \frac{7}{2}+1\times 4}{2+1},\frac{2\times \frac{9}{2}+1\times 2}{2+1})=(\frac{11}{3},\frac{11}{3})\\ \text{(iii)}\\ \text{Median BE of the triangle will divide the side AC in two equal parts.}\\ \text{Therefore, E is the mid-point of the side AC.}\\ \therefore \text{Coordinates of E}=(\frac{4+1}{2},\frac{2+4}{2})=(\frac{5}{2},3)\\ \text{Point Q divides the side BE in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of Q}=(\frac{2\times \frac{5}{2}+1\times 6}{2+1},\frac{2\times 3+1\times 5}{2+1})=(\frac{11}{3},\frac{11}{3})\\ \text{Median CF of the triangle will divide the side AB in two}\\ \text{equal parts.}\\ \text{Therefore, F is the mid-point of the side AB.}\\ \therefore \text{Coordinates of F}=(\frac{4+6}{2},\frac{2+5}{2})=(5,\frac{7}{2})\\ \text{Point R divides the side CF in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of R}=(\frac{2\times 5+1\times 1}{2+1},\frac{2\times \frac{7}{2}+1\times 4}{2+1})=(\frac{11}{3},\frac{11}{3})\\ \text{(iv)}\\ \text{We observe that coordinates of point P, Q and R are same.}\\ \text{Therefore, all these are representing the same point on}\\ \text{the plane i.e., the centroid of the triangle.}\\ \text{(v)}\\ \text{We consider}\mathrm{\Delta }\text{ABC having vertices A(}{\mathrm{x}}_1,{\mathrm{y}}_1\text{), B(}{\mathrm{x}}_2,{\mathrm{y}}_2\text{) and}\\ \text{C(}{\mathrm{x}}_3,{\mathrm{y}}_3\text{). Let AD is median and so D bisects side BC.}\\ \therefore \text{Coordinates of D}=(\frac{{\mathrm{x}}_2+{\mathrm{x}}_3}{2},\frac{{\mathrm{y}}_2+{\mathrm{y}}_3}{2})\\ \text{Let O is the centroid of}\mathrm{\Delta }\text{ABC}.\text{So, O divides median AD in}\\ \text{the ratio 2:1.}\\ \therefore \text{Coordinates of centroid}\mathrm{O}=(\frac{{\mathrm{x}}_1+2\left(\frac{{\mathrm{x}}_2+{\mathrm{x}}_3}{2}\right)}{2+1},\frac{{\mathrm{y}}_1+2\left(\frac{{\mathrm{y}}_2+{\mathrm{y}}_3}{2}\right)}{2+1})\\ =(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3})\end{array}$