Exercise 5.4 - Chapter 5 - Coordinate Geometry - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Question 1.
Find the coordinates of the point which divides the line segment joining the points $A(4,-3)$ and $B(9,7)$ in the ratio $3: 2$.
Solution:

By section formula $\mathrm{P}\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+m y_{1}}{m+n}\right)=\mathrm{P}(x, y)$
$\mathrm{P}(x, y)=\left(\frac{3(9)+2(4)}{3+2}, \frac{3(7)+2(-3)}{3+2}\right)=\left(\frac{27+8}{5}, \frac{21-6}{5}\right)=\left(\frac{35}{5}, \frac{15}{5}\right)=(7,3)$

Question $2 .$
In what ratio does the point $\mathrm{P}(2,-5)$ divide the line segment joining $\mathrm{A}(-3,5)$ and $\mathrm{B}(4,-9)$.
Solution:

$\begin{aligned}
\begin{array}{c}
x_{1} y_{1} \\
\mathrm{~A}(-3,5) \\
\mathrm{B}(4,-9)
\end{array} & \mathrm{P}(2,-5) \\
\mathrm{P}(x, y) &=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right) \\
\mathrm{P}(2,-5) &=\left(\frac{m \times 4+n(-3)}{m+n}, \frac{m(-9)+n(5)}{m+n}\right) \Rightarrow\left(\frac{4 m-3 n}{m+n}\right)=2 \\
4 m-3 n &=2 m+2 n \\
4 m-2 m &=2 n+3 n \\
2 m &=5 n \\
\frac{m}{n} &=\frac{5}{2} \\
\therefore \text { The ratio } m: n &=5: 2
\end{aligned}$
 

Question 3 .
Find the coordinates of a point $\mathrm{P}$ on the line segment joining $\mathrm{A}(1,2)$ and $\mathrm{B}(6,7)$ in such a way that $\mathrm{AP}=$ $\frac{2}{5} \mathrm{AB}$.

Solution:
$\begin{array}{cc}x_{1} y_{1} & x_{2} y_{2} \\ \mathrm{~A}(1,2) & \mathrm{B}(6,7) \\ \mathrm{P}(x, y)= & \left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)=\left(\frac{2 \times 6+3 \times 1}{2+3}, \frac{2 \times 7+3 \times 2)}{2+3}\right)\end{array}$
$=\left(\frac{12+3}{5}, \frac{14+6}{5}\right)=\left(\frac{15}{5}, \frac{20}{5}\right)=(3,4)$

Question $4 .$
Find the coordinates of the points of trisection of the line segment joining the points $A(-5,6)$ and $B(4,-3)$.

Solution:

P divides $\overline{\mathrm{AB}}$ in the ratio $m n$ $1: 2$ $\mathrm{P}(x, y)=\left(\frac{1 \times 4+2 \times-5}{1+2}, \frac{1 \times-3+2 \times 6}{1+2}\right)=\left(\frac{4-10}{3}, \frac{-3+12}{3}\right)=\left(\frac{-6}{3}, \frac{9}{3}\right)=(-2,3)$ $\mathrm{Q}(x, y)=\left(\frac{2 \times 4+1 \times-5}{2+1}, \frac{2 \times(-3)+1 \times 6}{2+1}\right)=\left(\frac{8-5}{3}, \frac{-6+6}{3}\right)=\left(\frac{3}{3}, \frac{0}{3}\right)=(1,0)$

Question $5 .$
The line segment joining $\mathrm{A}(6,3)$ and $\mathrm{B}(-1,-4)$ is doubled in length by adding half of $\mathrm{AB}$ to each end. Find the coordinates of the new end points.
Solution:

Question 6.
Using section formula, show that the points $A(7,-5), B(9,-3)$ and $C(13,1)$ are collinear.
Solution:
$\mathrm{A}(7,-5) \quad \frac{\mathrm{B}(9,-3)}{\mathrm{AB}}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=\sqrt{(9-7)^{2}+(-3-(-5))^{2}}$
$=\sqrt{2^{2}+(-3+5)^{2}}=\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}=2 \sqrt{2}$
$\overline{\mathrm{BC}}=\sqrt{(13-7)^{2}+(1-(-5))^{2}}=\sqrt{4^{2}+4^{2}}=\sqrt{16+16}$
$=\sqrt{32}=\sqrt{16 \times 2}=4 \sqrt{2}$
$\overline{\mathrm{AC}}=\sqrt{(13-7)^{2}+(1-(-5))^{2}}$
$=\sqrt{6^{2}+6^{2}}=\sqrt{36+36}=\sqrt{72}=\sqrt{2 \times 36}=6 \sqrt{2}$
$2 \sqrt{2}+4 \sqrt{2}=6 \sqrt{2}$
$\therefore \mathrm{AB}+\mathrm{BC}=\mathrm{AC}$,
Here B is the common Point. $\therefore \mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear

Question $7 .$
A line segment $\mathrm{AB}$ is increased along its length by $25 \%$ by producing it to $\mathrm{C}$ on the side of $\mathrm{B}$. If $\mathrm{A}$ and $\mathrm{B}$ have the coordinates $(-2,-3)$ and $(2,1)$ respectively, then find the coordinates of $\mathrm{C}$.
Solution:

$m: n=3: 1$
The point $\mathrm{P}$ divides $\mathrm{AB}$ in the ratio $3: 1$
$\mathrm{P}(x, y)=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)=\left(\frac{3 \times 2+1 \times-2}{3+1}, \frac{3 \times 1+1 \times-3)}{3+1}\right)$
$=\left(\frac{6-2}{4}, \frac{3-3}{4}\right)=\left(\frac{4}{4}, \frac{0}{4}\right)=(1,0)$
P is at $25 \%$ distance from $B$ on its left and $\mathrm{C}$ is at $25 \%$ distance from $B$ on its right.
$\therefore \mathrm{B}$ is the mid point of $\mathrm{PC}$
$\begin{aligned}&\text { Mid point of } \overline{\mathrm{PC}} \\&x_{1} y_{1} \quad \mathrm{C}\left(x_{2}, y_{2}\right)\end{aligned}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
$\begin{array}{rl}x_{1} y_{1} & \mathrm{C}\left(x_{2}, y_{2}\right) \\ \mathrm{P}(1,0) & \\ & (2,1)=\left(\frac{1+x_{2}}{2}, \frac{0+y_{2}}{2}\right)\end{array}$
$\frac{1+x_{2}}{2}=2 \quad \frac{y_{2}}{2}=1$
$1+x_{2}=4 \quad y_{2}=2$
$\Rightarrow \quad x_{2}=3, \quad y_{2}=2$
$\therefore \mathrm{C}\left(x_{2}, y_{2}\right)=(3,2)$ is the solutions.