NCERT Solutions for Exercise 7.2 Class 10 Maths Chapter 7 - Coordinate Geometry

Q2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Answer:

Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_1,y_1)$ and $Q(x_2,y_2)$

Then,

Section formula: $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

By observation point, P divides AB internally in the ratio $1:2$ .

Hence, $m:n=1:2$

Substituting the values in the equation we get;

$\Rightarrow P(\frac{1(-2)+2(4)}{1+2},\frac{1(-3)+2(-1)}{1+2})$

$\Rightarrow P(\frac{-2+8}{3},\frac{-3-2}{3})$

$\Rightarrow P(2,\frac{-5}{3})$

And by observation point Q, divides AB internally in the ratio $2:1$

Hence, $m:n=2:1$

Substituting the values in the equation above, we get

$\Rightarrow Q(\frac{2(-2)+1(4)}{2+1},\frac{2(-3)+1(-1)}{2+1})$

$\Rightarrow Q(\frac{-4+4}{3},\frac{-6-1}{3})$

$\Rightarrow Q(0,\frac{-7}{3})$

Hence, the points of trisections are $P(2,\frac{-5}{3})$ and $Q(0,\frac{-7}{3})$

Q3 To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag

Answer:

Niharika posted the green flag at the distance P, i.e.,

$\frac{1}{4}\times 100m=25m$ from the starting point of $2^{nd}$ line.

Therefore, the coordinates of this point $P$ are $(2,25).$

Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,

$\frac{1}{5}\times 100m=20m$ from the starting point of $8^{th}$ line.

Therefore, the coordinates of this point Q are $(8,20)$ .

The distance $PQ$ is given by,

$PQ=\sqrt{(8-2{)}^2+(25-20{)}^2}=\sqrt{36+25}=\sqrt{61}m$

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$ .

Then, by Section Formula,

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$x=\frac{2+8}{2},y=\frac{25+20}{2}$

$x=5,y=22.5$

Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.

Q4 Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Answer:

Let the ratio be : $k:1$

Then, By section formula:

$P(x,y)=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})$

Given point $P(x,y)=(-1,6)$

$-1=\frac{6k-3}{k+1}$

$\Rightarrow -k-1=6k-3$

$\Rightarrow k=\frac{7}{2}$

Hence, the point $P$ divides the line AB in the ratio $2:7$ .

Q5 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.

Answer:

Let the point on the x-axis be $P(x,0)$ and it divides it in the ratio $k:1$ .

Then, we have

Section formula:

$P(x,y)=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})$

$⟹\frac{k{y}_2+y_1}{k+1}=0$

$k=-\frac{y_1}{y_2}$

Hence, the value of k will be: $k=-\frac{-5}{5}=1$

Therefore, the x-axis divides the line in the ratio $1:1$ and the point will be,

Putting the value of $k=1$ in section formula.

$P(x,0)=(\frac{x_2+x_1}{2},0)$

$P(x,0)=(\frac{1-4}{2},0)=(\frac{-3}{2},0)$

Q6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:

Let the given points $A(1,2),B(4,y),C(x,6),D(3,5)$ .

Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

The coordinates of the point O when it is mid-point of AC.

$(\frac{1+x}{2},\frac{2+6}{2})\Rightarrow (\frac{x+1}{2},4)$

The coordinates of the point O when it is mid-point of BD.

$(\frac{4+3}{2},\frac{5+y}{2})\Rightarrow (\frac{7}{2},\frac{5+y}{2})$

Since both coordinates are of same point O.

Therefore,

$\frac{x+1}{2}=\frac{7}{2}$ and $4=\frac{5+y}{2}$

Or,

$x=6andy=3$

Q7 Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Answer:

As the centre point $C(2,-3)$ will be the mid-point of the diameter AB.

Then, the coordinates of point A will be $A(x,y)$ .

Given point $B(1,4)$ .

Therefore,

$(2,-3)=(\frac{x+1}{2},\frac{y+4}{2})$

$\frac{x+1}{2}=2and\frac{y+4}{2}=-3$

$\Rightarrow x=3andy=-10$ .

Therefore, the coordinates of A are $(3,-10).$

Q8 If A and B are (- 2, - 2) and (2, - 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Answer:

From the figure:

As $AP=\frac{3}{7}AB$

$\Rightarrow PB=\frac{4}{7}AB$ hence the ratio is 3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$P(x,y)=(\frac{3(2)+4(-2)}{3+4},\frac{3(-4)+4(-2)}{3+4})$

$P(x,y)=(\frac{6-8}{7},\frac{-12-8}{7})$

$P(x,y)=(\frac{-2}{7},\frac{-20}{7})$

Q9 Find the coordinates of the points which divide the line segment joining A (– 2, 2) and B(2, 8) into four equal parts.

Answer:

From the figure:

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Point C, D, and E.

Section Formula:

$P(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

Here point D divides the line segment AB into two equal parts hence

$D(x_2,y_2)=(\frac{-2+2}{2},\frac{2+8}{2})$

$D(x_2,y_2)=(0,5)$

Now, point C divides the line segment AD into two equal parts hence

$C(x_1,y_1)=(\frac{-2+0}{2},\frac{2+5}{2})$

$C(x_2,y_2)=(-1,\frac{7}{2})$

Also, point E divides the line segment DB into two equal parts hence

$E(x_1,y_1)=(\frac{2+0}{2},\frac{8+5}{2})$

$E(x_2,y_2)=(1,\frac{13}{2})$

Q10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer:

From the figure:

Let the vertices of the rhombus are:

$A(3,0),B(4,5),C(-1,4),D(-2,-1)$

Area of the rhombus ABCD is given by;

$=\frac{1}{2}\times (Productoflengthsofdiagonals)$

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Length of the diagonal AC:

$AC=\sqrt{(3-(-1){)}^2+(0-4{)}^2}=\sqrt{16+16}=4\sqrt{2}$

Length of the diagonal BD:

$BD=\sqrt{(4-(-2){)}^2+(5-(-1){)}^2}=\sqrt{36+36}=6\sqrt{2}$

Thus, the area will be,

$=\frac{1}{2}\times (AC)\times (BD)$

$=\frac{1}{2}\times (4\sqrt{2})\times (6\sqrt{2})=24squareunits.$