# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

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## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

In Coordinate Geometry, any point, figure, or object’s position is specified using two or more numbers. The location of the object can be specified in a variety of ways, including a line, a plane, three dimensions, etc. The Cartesian coordinate system is broadly used. It is one type of many that use a pair of numbers (coordinates), which are the signed distances between each point and two fixed perpendicular lines known as the axes, to represent each point. There are many different concepts in the chapter Coordinate Geometry which includes graphs, lines, etc. Therefore, students face difficulty in understanding the chapter clearly, as NCERT  textbooks do not contain detailed solutions. Students can refer to the Extramarks website for reliable and concise study material.

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## Access NCERT Solutions for Class-10 Maths Chapter 7 – Coordinate Geometry

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### NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

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Q.1

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

Ans

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

Q.2

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

Ans

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

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

Ans

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

Q.4 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,\text{}2\right)\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}\right).\\ \text{We know that sides of a square are equal to each other.}\\ \therefore \text{AB}=\text{BC}\\ \text{or}\sqrt{{\left(\mathrm{x}+1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}}=\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}}\\ \text{or}{\mathrm{x}}^{2}+2\mathrm{x}+1+{\left(\mathrm{y}-2\right)}^{2}={\mathrm{x}}^{2}-6\mathrm{x}+9+{\left(\mathrm{y}-2\right)}^{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}{\left(\mathrm{x}+1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}={\left(3+1\right)}^{2}+{\left(2-2\right)}^{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\right]\\ {\text{or 2y}}^{2}-8\mathrm{y}=0\\ {\text{or y}}^{2}-4\mathrm{y}=0\\ \text{or y(y}-4\right)=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}=\left(\frac{-1+3}{2},\text{}\frac{2+2}{2}\right)=\left(\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2},\text{}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2}\right)\\ \text{i.e.,}\left(\frac{\mathrm{x}+{\mathrm{x}}_{1}}{2},\text{}\frac{\mathrm{y}+{\mathrm{y}}_{1}}{2}\right)=\left(1,\text{}2\right)\\ \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.5 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}\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]\\ \text{}=\frac{1}{2}\left[4\left(2-5\right)+3\left(5-6\right)+6\left(6-2\right)\right]\\ \text{}=\frac{1}{2}\left[-12-3+24\right]\\ \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}\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}}=\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}\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]\\ \text{}=\frac{1}{2}\left[12\left(6-3\right)+13\left(3-2\right)+10\left(2-6\right)\right]\\ \text{}=\frac{1}{2}\left[36+13-40\right]\\ \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}\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}}=\frac{9}{2}\text{square units}\\ \text{We observe that area of the triangle in both cases is same.}\end{array}$

Q.6

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

Ans

$\begin{array}{l}\text{Given that,}\\ \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}}\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}=\left(\frac{1×1+3×4}{1+3},\text{}\frac{1×5+3×6}{1+3}\right)\\ \text{}=\left(\frac{13}{4},\text{}\frac{23}{4}\right)\\ \text{Coordinates of point E}=\left(\frac{1×7+3×4}{1+3},\text{}\frac{1×2+3×6}{1+3}\right)\\ \text{}=\left(\frac{19}{4},\text{}\frac{20}{4}\right)\\ \text{Area of a triangle}=\frac{1}{2}\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]\\ \text{Area of}\mathrm{\Delta }\text{ADE\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[4\left(\frac{23}{4}-\frac{20}{4}\right)+\frac{13}{4}\left(\frac{20}{4}-6\right)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]\\ \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}\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}}=\frac{1}{2}\left[3-\frac{13}{4}+\frac{19}{16}\right]\\ \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}\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}}=\frac{1}{2}\left[\frac{48-52+19}{16}\right]\\ \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}\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}}=\frac{15}{32}\text{square units}\\ \text{Area of}\mathrm{\Delta }\text{ABC\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[4\left(5-2\right)+1\left(2-6\right)+7\left(6-5\right)\right]\\ \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}\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}}=\frac{1}{2}\left[12-4+7\right]\\ \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}\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}}=\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.7

$\begin{array}{l}\text{Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of}\mathrm{\Delta }\text{ABC.}\\ \text{(i) The median from A meets BC at D. Find the}\\ \text{coordinates of the point D.}\\ \text{(ii) Find the coordinates of the point P on AD such that}\\ \text{AP}:\text{PD}=\text{2}:\text{1}\\ \text{(iii) Find the coordinates of points Q and R on medians}\\ \text{BE and CF respectively such that BQ}:\text{QE}=\text{2}:\text{1}\\ \text{and CR}:\text{RF}=\text{2}:\text{1.}\\ \text{(iv) What do yo observe?}\\ \text{[Note: The point which is common to all the three}\\ \text{medians is called the centroid and this point divides}\\ \text{each median in the ratio 2}:\text{1.]}\\ \text{(v) If A(}{\mathrm{x}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}\text{), B(}{\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}\text{) and C(}{\mathrm{x}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{3}\text{) are the vertices of}\\ \text{}\mathrm{\Delta }\text{ABC, find the coordinates of the centroid of the}\\ \text{triangle.}\end{array}$

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}=\left(\frac{6+1}{2},\text{}\frac{5+4}{2}\right)=\left(\frac{7}{2},\text{}\frac{9}{2}\right)\\ \text{(ii)}\\ \text{Point P divides the side AD in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of P}=\left(\frac{2×\frac{7}{2}+1×4}{2+1},\text{}\frac{2×\frac{9}{2}+1×2}{2+1}\right)=\left(\frac{11}{3},\text{}\frac{11}{3}\right)\\ \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}=\left(\frac{4+1}{2},\text{}\frac{2+4}{2}\right)=\left(\frac{5}{2},\text{}3\right)\\ \text{Point Q divides the side BE in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of Q}=\left(\frac{2×\frac{5}{2}+1×6}{2+1},\text{}\frac{2×3+1×5}{2+1}\right)=\left(\frac{11}{3},\text{}\frac{11}{3}\right)\\ \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}=\left(\frac{4+6}{2},\text{}\frac{2+5}{2}\right)=\left(5,\text{}\frac{7}{2}\right)\\ \text{Point R divides the side CF in the ratio 2}:\text{1.}\\ \therefore \text{Coordinates of R}=\left(\frac{2×5+1×1}{2+1},\text{}\frac{2×\frac{7}{2}+1×4}{2+1}\right)=\left(\frac{11}{3},\text{}\frac{11}{3}\right)\\ \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},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{1}\text{), B(}{\mathrm{x}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{2}\text{) and}\\ \text{C(}{\mathrm{x}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{3}\text{). Let AD is median and so D bisects side BC.}\\ \therefore \text{Coordinates of D}=\left(\frac{{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{2},\text{}\frac{{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{2}\right)\\ \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}=\left(\frac{{\mathrm{x}}_{1}+2\left(\frac{{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{2}\right)}{2+1},\text{}\frac{{\mathrm{y}}_{1}+2\left(\frac{{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{2}\right)}{2+1}\right)\\ \text{}=\left(\frac{{\mathrm{x}}_{1}+{\mathrm{x}}_{2}+{\mathrm{x}}_{3}}{3},\frac{{\mathrm{y}}_{1}+{\mathrm{y}}_{2}+{\mathrm{y}}_{3}}{3}\right)\end{array}$

Q.8 ABCD is a rectangle formed by the points A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Ans

$\begin{array}{l}\text{P is the mid-point of side AB.}\\ \therefore \text{Coordinates of P}=\left(\frac{-1-1}{2},\text{}\frac{-1+4}{2}\right)=\left(-1,\text{}\frac{3}{2}\right)\\ \text{Similarly, coordinates of Q, R and S are}\left(2,\text{}4\right),\text{}\left(5,\text{}\frac{3}{2}\right)\\ \text{and (2,}-1\right)\text{respectively.}\\ \text{Now, we find lengths of each side and each diagonal of}\\ \text{quadrilateral PQRS by using distance formula.}\\ \text{PQ}=\sqrt{{\left(-1-2\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{QR}=\sqrt{{\left(2-5\right)}^{2}+{\left(4-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{RS}=\sqrt{{\left(5-2\right)}^{2}+{\left(\frac{3}{2}+1\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{SP}=\sqrt{{\left(2+1\right)}^{2}+{\left(-1-\frac{3}{2}\right)}^{2}}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\\ \text{PR}=\sqrt{{\left(-1-5\right)}^{2}+{\left(\frac{3}{2}-\frac{3}{2}\right)}^{2}}=\sqrt{36+0}=6\\ \text{QS}=\sqrt{{\left(2-2\right)}^{2}+{\left(4+1\right)}^{2}}=\sqrt{0+25}=5\\ \text{We find that all sides of the quadrilateral PQRS are of equal}\\ \text{length, but diagonals are of different length. Therefore,}\\ \text{quadrilateral PQRS is a rhombus.}\end{array}$