Question

Find the image of the point $(-2,3)$ about the line $x-2y-9=0$.

Answer 1

Given a line $L_1=(x-2y-9)=0⟶\left(1\right)$

the line perpendicular to the above line and passing through $(-2,3)$ has a slope, say $M_1$

Slope of line $L_1=\frac{1}{2}$

Slope of the line perpendicular to $L_1=m_1=-2$ ($\because$ $m_1.m_2=-1$)

$\therefore$ the equation of the line perpendicular to $L_1$ and passing through $(-2,3)$ is given by

$y-3=-2(x+2)$

$\Rightarrow 2x+y+1=0⟶\left(2\right)$

Solving $\left(1\right)$ & $\left(2\right)$ simultaneously we will get the root of perpendicular drawn from the point $(-2,3)$ to the line $x-2y-9=0$

$\therefore$ Eqn $\left(1\right)\times 1⟶2x-2y-9=0$

Eqn $\left(2\right)\times 2⟶$$\underset{–}{4x+2y+2=0}$

(Adding) $5x-7=0$

$x=7/5$

$\therefore$ $y=-19/5$

the image of the point $(-2,3)$ about the line $x-2y-9=0$ will be at the same distance from $(\frac{7}{5},\frac{-19}{5})$ i.e, say the coordinates of image $ak(h,k)$

$\frac{7}{5}=\frac{-2+h}{2};\frac{19}{5}=\frac{3+k}{2}$

$\Rightarrow \frac{14}{5}+2=h;\frac{-38}{5}=3+k$

$\Rightarrow \frac{24}{5}=h;k=\frac{-53}{5}$

$\therefore$ the images is at $(\frac{24}{5},\frac{-53}{5})$