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

Q2 Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Answer:

If the points $(x, y), (1, 2)\ and\ (7, 0)$ are collinear then, the area formed by these points will be zero.

The area of the triangle is given by,

$Area = \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 0$

Substituting the values in the above equation, we have

$Area = \frac{1}{2}\left [ x(2-0)+1(0-y)+7(y-2) \right ]= 0$

$\Rightarrow 2x-y+7y-14= 0$

Or,

$\Rightarrow x+3y-7= 0$

Hence, the required relation between x and y is $x+3y-7= 0$ .

Q3 Find the center of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

Answer:

From the figure:

Let the center point be $O(x,y)$ .

Then the radii of the circle $OA,\ OB,\ and\ OC$ are equal.

The distance OA:

$OA = \sqrt{(x-6)^2+(y+6)^2}$

The distance OB:

$OB = \sqrt{(x-3)^2+(y+7)^2}$

The distance OC:

$OC = \sqrt{(x-3)^2+(y-3)^2}$

Equating the radii of the same circle.

When equating, $OA = OB$

$\sqrt{(x-6)^2+(y+6)^2}= \sqrt{(x-3)^2+(y+7)^2}$

Squaring both sides and applying $a^2-b^2 = (a+b)(a-b)$

$\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y+7)(y+6-y-7) = 0$

$\Rightarrow (2x-9)(-3) + (2y+13)(-1) = 0$

$\Rightarrow -6x+27-2y-13 = 0$ or

$\Rightarrow 3x+y -7= 0$ ...................................(1)

When equating, $OA = OC$

$\sqrt{(x-6)^2+(y+6)^2}= \sqrt{(x-3)^2+(y-3)^2}$

Squaring both sides and applying $a^2-b^2 = (a+b)(a-b)$

$\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y-3)(y+6-y+3) = 0$

$\Rightarrow (2x-9)(-3) + (2y+3)(9) = 0$

$\Rightarrow -3x+9y+27 = 0$ ...................................(2)

Now, adding the equations (1) and (2), we get

$\Rightarrow 10y = -20$

$\Rightarrow y = -2$ .

From equation (1), we get

$\Rightarrow 3x-2 = 7$

$\Rightarrow 3x =9$

$\Rightarrow x =3$

Therefore, the centre of the circle is $(3,-2)$ .

Q4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

Answer:

From the figure:

We know that the sides of a square are equal to each other.

Therefore, AB = BC

So,

$\sqrt{(x-1)^2+(y-2)^2} = \sqrt{(x-3)^2+(y-2)^2}$

Squaring both sides, we obtain

$\implies (x-1)^2+(y-2)^2 = (x-3)^2+(y-2)^2$

Now, doing $\left ( a^2-b^2 = (a+b)(a-b) \right )$

We get

$\implies (x-1+x-3)(x-1-x+3) = 0$

Hence $x = 2$ .

Applying the Pythagoras theorem to find out the value of y.

$AB^2+BC^2 = AC^2$

$(\sqrt{(2-1)^2+(y-2)^2})^2 + (\sqrt{(2-3)^2+(y-2)^2})^2 = (\sqrt{(3+1)^2+(2-2)^2})^2$

$\Rightarrow \left (\sqrt{1+(y-2)^2} \right )^2 + \left (\sqrt{1+(y-2)^2} \right )^2 = \left (\sqrt{16} \right )^2$

$\Rightarrow (y-2)^2 = 7$

Q5 (i) The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar is planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in Fig. 7.14. 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.

Answer:

Taking A as origin then, the coordinates of P, Q, and R can be found by observation:

Coordinates of point P is $(4,6).$

Coordinates of point Q is $(3,2).$

Coordinates of point R is $(6,5).$

The area of the triangle, in this case, will be:

$Area =\frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$=\frac{1}{2}\left [ 4(2-5)+3(5-6)+6(6-2) \right ]$

$=\frac{1}{2}\left [ -12-3+24\right ] = \frac{9}{2}\ Square\ units.$

Q5 (ii) The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar is planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in Fig. 7.14. The students are to sow seeds of flowering plants on the remaining area of the plot. (ii) What will be the coordinates of the vertices of D PQR if C is the origin? Also, calculate the areas of the triangles in these cases. What do you observe?

Answer:

Taking C as origin, then CB will be x-axis and CD be y-axis.

The coordinates fo the vertices P, Q, and R are: $(12,2),\ (13,6),\ (10,3).$ respectively.

The area of the triangle, in this case, will be:

$Area =\frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$=\frac{1}{2}\left [ 12(6-3)+13(3-2)+10(2-6) \right ]$

$=\frac{1}{2}\left [ 36-13+40 \right ] = \frac{9}{2}\ Square\ units.$

It can be observed that in both cases the area is the same so, it means that the area of any figure does not depend on the reference which you have taken.

Q6 The vertices of a $\Delta ABC$ are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that $\frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4}$ Calculate the area of the $\Delta ADE$ and compare it with the area of $\Delta ABC$ .

Answer:

From the figure:

Given ratio:

$\frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4}$

Therefore, D and E are two points on side AB and AC respectively, such that they divide side AB an AC in the ratio of $1:3$ .

Section formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Then, coordinates of point D:

$D(x_{1},y_{1})= \left (\frac{1\times1+3\times 4}{1+3} , \frac{1\times 5+3\times 6}{1+3} \right )$

Coordinates of point E:

$E(x_{2},y_{2})= \left (\frac{1\times7+3\times 4}{1+3} , \frac{1\times 2+3\times 6}{1+3} \right )$

$= \left ( \frac{19}{4}, \frac{20}{4} \right )$

Then, the area of a triangle:

$= \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

Substituting the values in the above equation,

$Area\ of\ \triangle ADE = \frac{1}{2}\left [ 4\left ( \frac{23}{4} - \frac{20}{4}\right )+\frac{13}{4}\left ( \frac{20}{4} - 6 \right )+\frac{19}{4}\left (6-\frac{23}{4} \right )\right ]$ $= \frac{1}{2}\left [ 3-\frac{13}{4} +\frac{19}{16}\right ] = \frac{1}{2}\left [ \frac{48-52+19}{16} \right ] = \frac{15}{32}\ square\ units.$

$Area\ of\ \triangle ABC = \frac{1}{2}\left [ 4(5-2)+1(2-6)+7(6-5) \right ]$

$= \frac{1}{2}\left [ 12-4+7 \right ] = \frac{15}{2}\ Square\ units.$

Hence the ratio between the areas of $\triangle ADE$ and $\triangle ABC$ is $1:16.$

Q7 (1) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of $\triangle ABC$ . The median from A meets BC at D. Find the coordinates of the point D.

Answer:

From the figure:

Let AD be the median of the triangle

Then, D is the mid-point of BC

Coordinates of Point D:

$\left ( \frac{6+1}{2},\frac{5+4}{2} \right ) = \left ( \frac{7}{2}, \frac{9}{2} \right )$

Q7 (ii) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. Find the coordinates of the point P on AD such that AP: PD = 2: 1

Answer:

From the figure,

The point P divides the median AD in the ratio, AP: PD = 2: 1

Hence using the section formula,

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$P(x,y)= \left (\frac{2\times\frac{7}{2}+1\times4}{2+1} , \frac{2\times\frac{9}{2}+1\times2}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

Q7 (iii) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. Find the coordinates of points Q and R on medians BE and CF respectively such that BQ: QE = 2: 1 and CR: RF = 2: 1

Answer:

From the figure,

$\Rightarrow$ The point Q divides the median BE in the ratio, BQ : QE = 2 : 1

Hence using the section formula,

$Q(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$Q(x,y)= \left (\frac{2\times\frac{5}{2}+1\times6}{2+1} , \frac{2\times3+1\times5}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

$\Rightarrow$ The point R divides the median CF in the ratio, CR: RF = 2: 1

Hence using the section formula,

$R(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$R(x,y)= \left (\frac{2\times 5+1\times1}{2+1} , \frac{2\times\frac{7}{2}+1\times4}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

Q7 (iv) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. What do you observe?

Answer:

We observed that the coordinates of P, Q, and R are the same. Therefore, all these are representing the same point on the plane. i.e., the centroid of the triangle.

Q7 (v) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of $\triangle ABC$ . If $A(x_1, y_1), B(x_2, y_2)\ and\ C(x_3, y_3)$ are the vertices of $\triangle ABC$ , find the coordinates of the centroid of the triangle.

Answer:

From the figure,

Let the median be AD which divides the side BC into two equal parts.

Therefore, D is the mid-point of side BC.

Coordinates of D:

$= \left ( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} \right )$

Let the centroid of this triangle be O.

Then, point O divides the side AD in a ratio 2:1.

Coordinates of O:

$= \left ( \frac{2\times\frac{x_{2}+x_{3}}{2}+1\times x_{1}}{2+1}, \frac{2\times\frac{y_{2}+y_{3}}{2}+1\times y_{1} }{2+1} \right )$

$= \left ( \frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3} \right )$

Q8 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.

Answer:

From the figure:

P is the mid-point of side AB.

Therefore, the coordinates of P are, $\left ( \frac{-1-1}{2}, \frac{-1+4}{2} \right ) = \left ( -1, \frac{3}{2} \right )$

Similarly, the coordinates of Q, R and S are: $\left ( 2,4 \right ),\ \left ( 5, \frac{3}{2} \right ),\ and\ \left ( 2,-1 \right )$ respectively.

The distance between the points P and Q:

$PQ = \sqrt{(-1-2)^2+\left ( \frac{3}{2} -4 \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

and the distance between the points Q and R:

$QR = \sqrt{(2-5)^2+\left ( 4-\frac{3}{2} \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points R and S:

$RS = \sqrt{(5-2)^2+\left ( \frac{3}{2}+1 \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points S and P:

$SP = \sqrt{(2+1)^2+\left ( -1-\frac{3}{2} \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points P and R the diagonal length:

$PR = \sqrt{(-1-5)^2+\left ( \frac{3}{2}-\frac{3}{2} \right )^2} = 6$

Distance between points Q and S the diagonal length:

$QS = \sqrt{(2-2)^2+\left ( 4+1 \right )^2} = 5$

Hence, it can be observed that all sides have equal lengths. However, the diagonals are of different lengths.

Therefore, PQRS is a rhombus.