Exercise 5.3 - Chapter 5 Coordinate Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.3$
Question 1.
Find the mid-points of the line segment joining the points
(i) $(-2,3)$ and $(-6,-5)$
(ii) $(8,-2)$ and $(-8,0)$
(iii) $(a, b)$ and $(a+2 b, 2 a-b)$
(iv) $\left(\frac{1}{2},-\frac{2}{7}\right)$ and $\left(\frac{3}{2}, \frac{-11}{7}\right)$
Solution:
(i) $(-2,3)$ and $(-6,-5)$
$\text { A }(-2,3), B(-6,-5)$
Hint : Mid point $\mathrm{M}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{(-2)+(-6)}{2}, \frac{3+(-5)}{2}\right)=\left(\frac{-8}{2}, \frac{-2}{2}\right)=(-4,-1)$
(ii) $\quad(8,-2)$ and $(-8,0)$
Hint : A $(8,-2), B(-8,0)$
Mid point $\mathrm{M}(x, y)=\left(\frac{8+(-8)}{2}, \frac{-2+0}{2}\right)=\left(\frac{0}{2}, \frac{-2}{2}\right)=(0,-1)$
(iii) $(a, b)$ and $(a+2 b, 2 a-b)$
Hint : A $(a, b), \mathrm{B}(a+2 b, 2 a-b)$
Mid point $\mathrm{M}(x, y)=\left(\frac{a+a+2 b}{2}, \frac{\not b+2 a-\not b}{2}\right)=\left(\frac{2 a+2 b}{2}, \frac{\not a a}{\not 2}\right)=\left(\frac{\not(a+b)}{\not{z}}, a\right)=(a+b, a)$
(iv) $\left(\frac{1}{2}, \frac{-3}{7}\right)$ and $\left(\frac{3}{2}, \frac{-11}{7}\right)$
Hint : $\mathrm{A}\left(\frac{1}{2}, \frac{-3}{7}\right), \mathrm{B}\left(\frac{3}{2}, \frac{-11}{7}\right)$
mid point $\mathrm{M}(x, y)=\left(\frac{\frac{1}{2}+\frac{3}{2}}{2}, \frac{\frac{-3}{7}+\frac{(-11)}{7}}{2}\right)\left(\frac{\frac{4^{2}}{\frac{1}{2}}}{2}-\frac{\frac{8^{1}}{\frac{1}{x}}}{8}\right)=(1,-1)$

Question $2 .$
The centre of a circle is $(-4,2)$. If one end of the diameter of the circle is $(-3,7)$ then find the other end.
Solution:

$\begin{aligned}
\mathrm{M}(x, y) &=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\
(-4,2) &=\left(\frac{-3+x_{2}}{2}, \frac{7+y_{2}}{2}\right) \\
\frac{-3+x_{2}}{2} &=-4 \quad \frac{7+y_{2}}{2}=2 \\
-3+x_{2} &=-8 \quad(x\\
x &=-8+3 \quad y_{2}=4 \\
x_{2} &=-5 \\
\text {. The other end is }(-5,-3) & y_{2}=4-7=-3
\end{aligned}$

Question $3 .$
If the mid-point $(x, y)$ of the line joining $(3,4)$ and $(p, 7)$ lies on $2 x+2 y+1=0$, then what will be the value of $p$ ?
Solution:

$\begin{aligned}
\mathrm{M}(x, y) &=\left(\frac{3+p}{2}, \frac{4+7}{2}\right) \\
\frac{3+p}{2} &=x \\
y &=\frac{11}{2}
\end{aligned}$
substitute the values of $(x, y)$ in $2 x+2 y+1=0$
$\begin{aligned}
2\left(\frac{3+p}{2}\right)+2\left(\frac{11}{2}\right)+1 &=0 \\
3+p+11+1 &=0 \\
p+15 &=0 \\
p &=-15
\end{aligned}$

Question $4 .$
The midpoint of the sides of a triangle are $(2,4),(-2,3)$ and $(5,2)$. Find the coordinates of the vertices of the triangle.
Solution:
Mid point
$\begin{aligned}
\mathrm{M}(x, y) &=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\
\text { Mid point } \mathrm{AB}(2,4) &=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\
\frac{x_{1}+x_{2}}{2} &=2 \Rightarrow x_{1}+x_{2}=4
\end{aligned}$

$\begin{aligned}
\frac{y_{1}+y_{2}}{2} &=4 \Rightarrow y_{1}+y_{2}=8 \\
\text { Mid point of } \mathrm{BC}(-2,3) &=\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right) \\
\left(\frac{x_{2}+x_{3}}{2}\right) &=-2 \Rightarrow x_{2}+x_{3}=-4 \\
\frac{y_{2}+y_{3}}{2} &=3 \Rightarrow y_{2}+y_{3}=6 \\
\text { Mid point of AC }(5,2) &=\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right) \\
\left(\frac{x_{1}+x_{3}}{2}\right) &=5 \Rightarrow x_{1}+x_{3}=10 \\
\frac{y_{1}+y_{3}}{2} &=2 \Rightarrow y_{1}+y_{3}=4
\end{aligned}$

Question $5 .$
$\mathrm{O}(0,0)$ is the centre of a circle whose one chord is $\mathrm{AB}$, where the points $\mathrm{A}$ and $\mathrm{B}$ are $(8,6)$ and $(10,0)$ respectively. $\mathrm{OD}$ is the perpendicular from the centre to the chord $\mathrm{AB}$. Find the coordinates of the mid-point of OD.
Solution:
Mid point $=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
Mid point of $\mathrm{AB}=\left(\frac{8+10}{2}, \frac{6+0}{2}\right)=\left(\frac{18}{2}, \frac{6}{2}\right)=(9,3)$
Mid point $\mathrm{M}(x, y)=$ Mid point of $\mathrm{OD}$
$=\left(\frac{0+9}{2}, \frac{0+3}{2}\right)=\left(\frac{9}{2}, \frac{3}{2}\right)=(4.5,1.5)$
 

Question 6.
The points $\mathrm{A}(-5,4), \mathrm{B}(-1,-2)$ and $\mathrm{C}(5,2)$ are the vertices of an isosceles right angled triangle where the right angle is at $\mathrm{B}$. Find the coordinates of $D$ so that $A B C D$ is a square.
Solution:
In squares the diagonals are equal and bisect each other $\therefore$ Mid point of $\mathrm{BD}=$ Mid point of $\mathrm{AC}$
$\left(\frac{-1+x}{2}, \frac{-2+y}{2}\right)=\left(\frac{-5+5}{2}, \frac{4+2}{2}\right)$
$\frac{-1+x}{2}=\frac{0}{2} \quad \frac{-2+y}{2}=\frac{6}{2}$
$-1+x=0$
$-2+y=6$
$y=8$
$\therefore$ The vertex $\mathrm{D}(x, y)=(1,8)$

Question 7.
The points $\mathrm{A}(-3,6), \mathrm{B}(0,7)$ and $\mathrm{C}(1,9)$ are the mid points of the sides $\mathrm{DE}, \mathrm{EF}$ and $\mathrm{FD}$ of a triangle DEF. Show that the quadrilateral $\mathrm{ABCD}$ is a parallelogram.
Solution:
In a parallelogram diagonals biscct each other and diagonals are not equal.
$\begin{aligned}
&\text { Mid point of } \mathrm{DE}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\
&\begin{array}{l}
(-3,6)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\
\frac{x_{1}+x_{2}}{2}=-3 \\
\frac{y_{1}+y_{2}}{2}=6 \\
x_{1}+x_{2}=-6 \quad \ldots y_{2}=12 \quad \ldots(1) \quad(2)
\end{array}
\end{aligned}$
Mid point of EF $=\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$
$(0,7)=\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$
$\frac{x_{2}+x_{3}}{2}=0 \quad \frac{y_{2}+y_{3}}{2}=7$
Mid points of the diagonals are equal in parallelogram
$\therefore$ We have to prove this
Mid point of $\mathrm{AC}=\left(\frac{(-3)+(-3)}{2}, \frac{6+(-2)}{2}\right)=\left(\frac{-6}{2}, \frac{4}{2}\right)=(-3,2)$
Mid point of $\mathrm{BD}=\left(\frac{-6+0}{2}, \frac{-3+7}{2}\right)=\left(\frac{-6}{2}, \frac{4}{2}\right)=(-3,2)$
$\therefore$ Mid point of $A C=$ Mid point of $B D$
$\therefore \mathrm{ABCD}$ is a parallelogram

Question $8 .$
A $(-3,2), \mathbf{B}(3,2)$ and $\mathrm{C}(-3,-2)$ are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.
Solution:

Mid point of $B C$
$=\left(\frac{3+(-3)}{2}, \frac{2+(-2)}{2}\right)=\left(\frac{0}{2}, \frac{0}{2}\right)=(0,0)$
Distance between two points
$\begin{aligned}
d &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\
\overline{\mathrm{OA}} &=\sqrt{(-3-0)^{2}+(2-0)^{2}} \\
&=\sqrt{(-3)^{2}+2^{2}}=\sqrt{9+4}=\sqrt{13} \\
\overline{\mathrm{OB}} &=\sqrt{(-3-0)^{2}+(2-0)^{2}}=\sqrt{9+4}=\sqrt{13} \\
\overline{\mathrm{OC}} &=\sqrt{(-3-0)^{2}+(-2-0)^{2}}=\sqrt{9+4}=\sqrt{13} \\
& \therefore \overline{\mathrm{OA}}=\overline{\mathrm{OB}}=\overline{\mathrm{OC}} .
\end{aligned}$
Hence proved