Exercise 5.5 - Chapter 5 - Coordinate Geometry - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 5.5 - Chapter 5 - Coordinate Geometry - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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 $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)=\left(\frac{\frac{3}{2}+\frac{7}{1}+\frac{13}{2}}{3}, \frac{5+\left(\frac{-9}{2}\right)+\frac{(-13)}{2}}{3}\right)$
$=\left(\frac{\frac{3+14+13}{2}}{3}, \frac{\frac{10-9-13}{2}}{3}\right)=\left(\frac{\frac{30}{2}}{3}, \frac{\frac{-12}{2}}{3}\right)=\left(\frac{15}{3}, \frac{-6}{3}\right)=(5,-2)$