Exercise 5.5 - Chapter 5 Coordinate Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 5.5
Question 1.
Find the centroid of the triangle whose vertices are
(i) $(2,-4),(-3,-7)$ and $(7,2)$
(ii) $(-5,-5),(1,-4)$ and $(-4,-2)$
Solution:
$\begin{array}{crc}x_{1} y_{1} & x_{2} y_{2} & x_{3} y_{3} \\ (2,-4) & (-3,-7) & (7,2)\end{array}$
(i) Centroid $\mathrm{G}(x, y)=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$
$=\left(\frac{(2)+(-3)+7}{3}, \frac{(-4)+(-7)+(2)}{3}\right)=\left(\frac{6}{3}, \frac{-9}{3}\right)=(2,-3)$
(ii) $x_{1} y_{1} \quad x_{2} y_{2} \quad x_{3} y_{3}$
$(2,-4) \quad(1,-4) \quad(-4,-2)$
Centroid $\mathrm{G}(x, y)=\left(\frac{(-5)+1+(-4)}{2}, \frac{(-5)+(-4)+(-2)}{3}\right)=\left(\frac{-8}{3}, \frac{-11}{3}\right)$
 

Question 2.
If the centroid of a triangle is at $(4,-2)$ and two of its vertices are $(3,-2)$ and $(5,2)$ then find the third vertex of the triangle.
Solution:

Centroid $\mathrm{G}(\mathrm{x}, \mathrm{y})=(4,-2)$
two vertices
$\left(x_{1}, y_{1}\right)=(3,-2)$
$\begin{aligned}
\left(x_{2}, y_{2}\right) &=(5,2),\left(x_{3}, y_{3}\right)=? \\
(4,-2)=\left(\frac{3+5+x_{3}}{3}, \frac{x_{2}+2+y_{3}}{3}\right)
\end{aligned}$
$(4,-2)=\left(\frac{8+x_{3}}{3}, \frac{y_{3}}{3}\right)$
$\begin{aligned}
\frac{8+x_{3}}{3} &=4 \quad \frac{y_{3}}{3}=-2 \\
8+x_{3} &=12 y_{3}=-6 \\
x_{3} &=4
\end{aligned}$
$\therefore$ The third vertex $\left(x_{3}, y_{3}\right)=(4,-6)$

Question $3 .$
Find the length of median through Aof a triangle whose vertices are $\mathrm{A}(-1,3), \mathrm{B}(1,-1)$ and $\mathrm{C}(5,1)$. Solution:
$\mathrm{D}(x, y)$ is the Mid point $\mathrm{BC}$
$\therefore \mathrm{D}(x, y)=\left(\frac{1+5}{2}, \frac{-1+1}{2}\right)=\left(\frac{6}{2}, \frac{0}{2}\right)=(3,0)$
$\mathrm{AD}$ is the median through $\mathrm{A}$
Here A $\quad x_{2} y_{2} \mathrm{D} \quad x_{3} y_{3}$
$\therefore$ Length of AD $\quad(-1,3) \quad \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=\sqrt{(3-(-1))^{2}+(0-3)^{2}}$
$=\sqrt{(3+1)^{2}+(-3)^{2}}=\sqrt{16+9}=\sqrt{25}=5$ units

Question $4 .$
The vertices of a triangle are $(1,2),(\mathrm{h},-3)$ and $(-4, \mathrm{k})$. If the centroid of the triangle is at the point $(5,-1)$ then find the value of $\sqrt{(h+k)^{2}+(h+3 k)^{2}}$

Solution:
Vertices of a triangle
$\begin{aligned}
&\left(x_{1}, y_{1}\right)=(1,2) \\
&\left(x_{2}, y_{2}\right)=(h,-3) \\
&\left(x_{3}, y_{3}\right)=(-4, k) \\
&\text { Centroid } \mathrm{G}(x, y)=(5,-1) \\
&\mathrm{G}= \\
&\qquad\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)=(5,-1) \\
&\Rightarrow \quad \frac{-3+h}{3}=5 \frac{2-3+k}{3} \mid=-1 \\
&\quad-3+h=15-1+k=-3 \mid \\
&h=18 \\
&\left.\quad \therefore \sqrt{(h+k)^{2}+(h+3 k)^{2}}\right)=(5,-1) \\
&\quad=\sqrt{(18+(-2))^{2}+(18+3(-2))^{2}}=\sqrt{16^{2}+12^{2}}=\sqrt{256+144}=\sqrt{400}=20
\end{aligned}$

Question $5 .$
Orthocentre and centroid of a triangle are $\mathrm{A}(-3,5)$ and $\mathrm{B}(3,3)$ respectively. If $\mathrm{C}$ is the circumcentre and $\mathrm{AC}$ is the diameter of this circle, then find the radius of the circle.
Solution:

$\begin{aligned}
&\mathrm{H}=\mathrm{A}(-3,5)\\
&\mathrm{G}=\mathrm{B}(3,3)\\
&\mathrm{S}=\text { Circumcentre (C) }\\
&\mathrm{G} \text { divides } \mathrm{H} \text { and } \mathrm{S} \text { in the ratio } 2: 1\\
&m: n=1: 2\\
&\mathrm{A}\left(x_{1}, y_{1}\right)=(-3,5)\\
&\begin{aligned}
&\mathrm{C}\left(x_{2}, y_{2}\right)=\left(x_{2}, y_{2}\right) \\
&\mathrm{P}(x, y)=(3,3)
\end{aligned}\\
&\mathrm{P}(x, y)=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\\
&(3,3)=\left(\frac{1 x_{2}+2(3)}{2+1}, \frac{1\left(y_{2}\right)+2(5)}{2+1}\right)\\
&(3,3)=\left(\frac{x_{2}-6}{3}, \frac{y_{2}+10}{3}\right)\\
&\frac{x_{2}-6}{3}=3 \quad \frac{y_{2}+10}{3}=3\\
&x_{2}-6=9 \quad y_{2}+10=9\\
&x_{2}=9+6 \quad y_{2}=9-10\\
&x_{2}=15 \quad y_{2}=-1\\
&\overline{\mathrm{AC}}=\sqrt{(5-(-3))^{2}+(-1-5)^{2}}=\sqrt{8^{2}+(-6)^{2}}=\sqrt{324+36}\\
&=\sqrt{360}=\sqrt{36 \times 10}=6 \sqrt{10}\\
&\text { radius }=\frac{1}{2} \mathrm{AC} \quad \because \mathrm{AC} \text { is a diametre of circle }\\
&=\frac{1}{2} \times 6 \sqrt{10}=3 \sqrt{10} \text { units. }
\end{aligned}$

Question 6 .
$\mathrm{ABC}$ is a triangle whose vertices are $\mathrm{A}(3,4), \mathrm{B}(-2,-1)$ and $\mathrm{C}(5,3)$. If $\mathrm{G}$ is the centroid and $\mathrm{BDCG}$ is a parallelogram then find the coordinates of the vertex $D$.
Solution:
$\begin{aligned}
\text { Centroid } \mathrm{G} &=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right) \\
\therefore \mathrm{G}(x, y) &=\left(\frac{3+(-2)+5}{3}, \frac{4+(-1)+3}{3}\right) \\
&=\left(\frac{8-2}{3}, \frac{7-1}{3}\right)=\left(\frac{6}{3}, \frac{6}{3}\right)=(2,2)
\end{aligned}$
In a parallelogram diagonals bisect each other
$\therefore$ Mid point of $D G=$ Mid point of $B C$
$\left(\frac{x+2}{2}, \frac{y+2}{2}\right)=\left(\frac{-2+5}{2}, \frac{-1+3}{2}\right)$
$\begin{array}{rlrl}
\frac{x+2}{2} & =\frac{3}{2} & \frac{y+2}{2} & =\frac{2}{2} \\
x+2 & =3 & y & =2-2=0 \\
x & =3-2=1 & &
\end{array}$
$\therefore$ The co-ordinates of the vertex $D(x, y)=(1,0)$

Question 7.
If and are mid points of the sides of a triangle, then find the centroid of the triangle.
Solution:
"The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle."
$\therefore$ The mid points of the sides of the triangle are given as
$\therefore$ Centroid $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)=\left(\frac{\frac{3}{2}+\frac{7}{1}+\frac{13}{2}}{3}, \frac{5+\left(\frac{-9}{2}\right)+\frac{(-13)}{2}}{3}\right)$
$=\left(\frac{\frac{3+14+13}{2}}{3}, \frac{\frac{10-9-13}{2}}{3}\right)=\left(\frac{\frac{30}{2}}{3}, \frac{\frac{-12}{2}}{3}\right)=\left(\frac{15}{3}, \frac{-6}{3}\right)=(5,-2)$