Exercise 7.2 (Revised) - Chapter 7 - Coordinate Geometry - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths: Chapter 7 Coordinate Geometry Solutions

Ex 7.2 Question 1.

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

Answer.

Let $x_1=-1, x_2=4, y_1=7$ and $y_2=-3, m_1=2$ and $m_2=3$
Using Section Formula to find coordinates of point which divides join of $(-1,7)$ and $(4,-3)$ in the ratio $2: 3$, we get
$
\begin{aligned}
& x=\frac{m_1 x_2+m_2 x_1}{m_1+m_2}=\frac{2 \times 4+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1 \\
& y=\frac{m_1 y_2+m_2 y_1}{m_1+m_2}=\frac{2 \times(-3)+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3
\end{aligned}
$

Therefore, the coordinates of point are $(1,3)$ which divides join of $(-1,7)$ and $(4,-3)$ in the ratio 2:3.

Ex 7.2 Question 2.

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

Answer

We want to find coordinates of the points of trisection of the line segment joining $(4,-1)$ and $(-2,-3)$.
We are given $\mathrm{AC}=\mathrm{CD}=\mathrm{DB}$
We want to find coordinates of point $C$ and $D$.
Let coordinates of point $\mathrm{C}$ be $\left(x_1, y_1\right)$ and let coordinates of point $\mathrm{D}$ be $\left(x_2, y_2\right)$.
Clearly, point $\mathrm{C}$ divides line segment $\mathrm{AB}$ in $1: 2$ and point $\mathrm{D}$ divides line segment $\mathrm{AB}$ in 2:1.
Using Section Formula to find coordinates of point $C$ which divides join of $(4,-1)$ and $(-2,-3)$

in the ratio $1: 2$, we get
$
\begin{aligned}
& x_1=\frac{1 \times(-2)+2 \times 4}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2 \\
& y_1=\frac{1 \times(-3)+2 \times(-1)}{1+2}=\frac{-3-2}{3}=\frac{-5}{3}
\end{aligned}
$

Using Section Formula to find coordinates of point D which divides join of $(4,-1)$ and $(-2,-3)$ in the ratio $2: 1$, we get
$
\begin{aligned}
& x_2=\frac{2 \times(-2)+1 \times 4}{1+2}=\frac{-4+4}{3}=\frac{0}{3}=0 \\
& y_2=\frac{2 \times(-3)+1 \times(-1)}{1+2}=\frac{-6-1}{3}=\frac{-7}{3}
\end{aligned}
$

Therefore, coordinates of point $\mathrm{C}$ are $\left(2,-\frac{5}{3}\right)$ and coordinates of point $\mathrm{D}$ are $\left(0,-\frac{7}{3}\right)$
Ex 7.2 Question 3.

To conduct sports day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of $1 \mathrm{~m}$ each. 100 flower pots have been placed at a distance of $1 \mathrm{~m}$ from each other along $A D$. Niharika runs 14th of the distance $A D$ on the 2nd line and posts a green flag. Preet runs 15th of 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 runs $14^{\text {th }}$ of the distance $A D$ on the $2^{\text {nd }}$ line and posts a green flag.
There are 100 flower pots. It means, she stops at 25 th flower pot.
Therefore, the coordinates of point where she stops are ( $2 \mathrm{~m}, 25 \mathrm{~m})$.
Preet runs 15th of the distance $\mathrm{AD}$ on the eighth line and posts a red flag. There are 100 flower pots. It means, she stops at 20th flower pot.

Therefore, the coordinates of point where she stops are $(8,20)$.
Using Distance Formula to find distance between points $(2 \mathrm{~m}, 25 \mathrm{~m})$ and $(8 \mathrm{~m}, 20 \mathrm{~m})$, we get
$
d=\sqrt{(2-8)^2+(25-20)^2}=\sqrt{(-6)^2+5^2}=\sqrt{36+25}=\sqrt{61} m
$

Rashmi posts a blue flag exactly halfway the line segment joining the two flags.
Using section formula to find the coordinates of this point, we get
$
\begin{aligned}
& x=\frac{2+8}{2}=\frac{10}{2}=5 \\
& y=\frac{25+20}{2}=\frac{45}{2}
\end{aligned}
$

Therefore, coordinates of point, where Rashmi posts her flag are $\left(5, \frac{45}{2}\right)\left(5, \frac{45}{2}\right)$.
It means she posts her flag in 5th line after covering $\frac{45}{2}=22.5 \mathrm{~m}$ of distance.

Ex 7.2 Question 4.

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

Answer.

Let $(-1,6)$ divides line segment joining the points $(-3,10)$ and $(6,-8)$ in $\mathrm{k}: 1$.
Using Section formula, we get
$
-1=\frac{(-3) \times 1+6 \times k}{k+1}
$

$
\begin{aligned}
& \Rightarrow-\mathrm{k}-1=(-3+6 \mathrm{k}) \\
& \Rightarrow-7 \mathrm{k}=-2 \\
& \Rightarrow \mathrm{k}=\frac{2}{7}
\end{aligned}
$

Therefore, the ratio is $\frac{2}{7}: 1$ which is equivalent to $2: 7$.
Therefore, $(-1,6)$ divides line segment joining the points $(-3,10)$ and $(6,-8)$ in 2:7.
Ex 7.2 Question 5.

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 coordinates of point of division be ( $\mathrm{x}, 0$ ) and suppose it divides line segment joining A $(1,-5)$ and $B(-4,5)$ in $k: 1$.

According to Section formula, we get
$
\begin{aligned}
& x=\frac{1 \times 1+(-4) \times k}{k+1}=\frac{1-4 k}{k+1} \text { and } 0=\frac{(-5) \times 1+5 k}{k+1} \ldots \\
& 0=\frac{(-5) \times 1+5 k}{k+1} \\
& \Rightarrow 5=5 \mathrm{k} \\
& \Rightarrow \mathrm{k}=1
\end{aligned}
$

Putting value of $\mathrm{k}$ in (1), we get
$
x=\frac{1 \times 1+(-4) \times 1}{1+1}=\frac{1-4}{2}=\frac{-3}{2}
$

Therefore, point $\left(\frac{-3}{2}, 0\right)$ on $x$-axis divides line segment joining $A(1,-5)$ and $B(-4,5)$ in 1:1.

Ex 7.2 Question 6.

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

Answer.

Let $\mathrm{A}=(1,2), \mathrm{B}=(4, \mathrm{y}), \mathrm{C}=(\mathrm{x}, 6)$ and $\mathrm{D}=(3,5)$
We know that diagonals of parallelogram bisect each other. It means that coordinates of midpoint of diagonal $A C$ would be same as coordinates of midpoint of diagonal BD.

Using Section formula, the coordinates of midpoint of $\mathrm{AC}$ are:
$
\frac{1+x}{2}, \frac{2+6}{2}=\frac{1+x}{2}, 4
$

Using Section formula, the coordinates of midpoint of $\mathrm{BD}$ are:
$
\frac{4+3}{2}, \frac{5+y}{2}=\frac{7}{2}, \frac{5+y}{2}
$

According to condition (1), we have
$
\begin{aligned}
& \frac{1+x}{2}=\frac{7}{2} \\
& \Rightarrow(1+x)=7 \\
& \Rightarrow x=6
\end{aligned}
$

Again, according to condition (1), we also have
$
\begin{aligned}
& 4=\frac{5+y}{2} \\
& \Rightarrow 8=5+y \\
& \Rightarrow y=3
\end{aligned}
$

Therefore, $x=6$ and $y=3$

Ex 7.2 Question 7.

Find the coordinates of a point $\mathrm{A}$, where $\mathrm{AB}$ is the diameter of a circle whose centre is $(2,-3)$ and $B$ is $(1,4)$.

Answer.

We want to find coordinates of point A. AB is the diameter and coordinates of center

are $(2,-3)$ and, coordinates of point $B$ are $(1,4)$.
Let coordinates of point A are (x, y). Using section formula, we get
$
\begin{aligned}
& 2=\frac{x+1}{2} \\
& \Rightarrow 4=x+1 \\
& \Rightarrow x=3
\end{aligned}
$

Using section formula, we get
$
\begin{aligned}
& -3=\frac{4+y}{2} \\
& \Rightarrow-6=4+y \\
& \Rightarrow y=-10
\end{aligned}
$

Therefore, Coordinates of point A are $(3,-10)$.
Ex 7.2 Question 8.

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

Answer.

$A=(-2,-2)$ and $B=(2,-4)$

It is given that $\mathrm{AP}=\frac{3}{7} \mathrm{AB}$
$
\mathrm{PB}=\mathrm{AB}-\mathrm{AP}=\mathrm{AB}-\frac{3}{7} \mathrm{AB}=\frac{4}{7} \mathrm{AB}
$

So, we have AP: $\mathrm{PB}=3: 4$
Let coordinates of $\mathrm{P}$ be $(\mathrm{x}, \mathrm{y})$

Using Section formula to find coordinates of $\mathrm{P}$, we get
$
\begin{aligned}
& x=\frac{(-2) \times 4+2 \times 3}{3+4}=\frac{6-8}{7}=\frac{-2}{7} \\
& y=\frac{(-2) \times 4+(-4) \times 3}{3+4}=\frac{-8-12}{7}=\frac{-20}{7}
\end{aligned}
$

Therefore, Coordinates of point $\mathrm{P}$ are $\left(\frac{-2}{7}, \frac{-20}{7}\right)$.
Ex 7.2 Question 9.

Find the coordinates of the points which divides the line segment joining $A(-2,2)$ and B $(2,8)$ into four equal parts.

Answer.

$A=(-2,2)$ and $B=(2,8)$

Let $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are the points which divide line segment $\mathrm{AB}$ into 4 equal parts.
Let coordinates of point $P=\left(x_1, y_1\right), Q=\left(x_2, y_2\right)$ and $R=\left(x_3, y_3\right)$
We know $\mathrm{AP}=\mathrm{PQ}=\mathrm{QR}=\mathrm{RS}$.
It means, point $P$ divides line segment $A B$ in $1: 3$.
Using Section formula to find coordinates of point $\mathrm{P}$, we get
$
\begin{aligned}
& x_1=\frac{(-2) \times 3+2 \times 1}{1+3}=\frac{-6+2}{4}=\frac{-4}{4}=-1 \\
& y_1=\frac{2 \times 3+8 \times 1}{1+3}=\frac{6+8}{4}=\frac{14}{4}=\frac{7}{2}
\end{aligned}
$

Since, $A P=P Q=Q R=R S$.
It means, point $Q$ is the mid-point of $A B$.
Using Section formula to find coordinates of point $Q$, we get
$
\begin{aligned}
& x_2=\frac{(-2) \times 1+2 \times 1}{1+1}=\frac{-2+2}{2}=\frac{0}{2}=0 \\
& y_2=\frac{2 \times 1+8 \times 1}{1+1}=\frac{2+8}{2}=\frac{10}{2}=5
\end{aligned}
$

Because, $\mathrm{AP}=\mathrm{PQ}=\mathrm{QR}=\mathrm{RS}$.
It means, point $\mathrm{R}$ divides line segment $\mathrm{AB}$ in $3: 1$
Using Section formula to find coordinates of point $P$, we get
$
\begin{aligned}
& x_3=\frac{(-2) \times 1+2 \times 3}{1+3}=\frac{-2+6}{4}=\frac{4}{4}=1 \\
& y_3=\frac{2 \times 1+8 \times 3}{1+3}=\frac{2+24}{4}=\frac{26}{4}=\frac{13}{2}
\end{aligned}
$

Therefore, $\mathrm{P}=\left(-1, \frac{7}{2}\right), \mathrm{Q}=(0,5)$ and $\mathrm{R}=\left(1, \frac{13}{2}\right)$
Ex 7.2 Question 10.

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

Answer.

Let $\mathrm{A}=(3,0), \mathrm{B}=(4,5), \mathrm{C}=(-1,4)$ and $\mathrm{D}=(-2,-1)$
Using Distance Formula to find length of diagonal $\mathrm{AC}$, we get
$
A C=\sqrt{[3-(-1)]^2+(0-4)^2}=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}
$

Using Distance Formula to find length of diagonal $\mathrm{BD}$, we get
$
B D=\sqrt{[4-(-2)]^2+[5-(-1)]^2}=\sqrt{6^2+6^2}=\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}
$
$\because$ Area of rhombus $=1 / 2$ (product of its diagonals)
$
=\frac{1}{2} \times A C \times B D=\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}=24 \text { sq. units }
$