Question

If $(-2,6)$ is the image of point $(4,2)$ with respect to the line $L=0$, then equation of line is

• A. $6x-4y-7=0$
• B. $2x+3y-5=0$
• C. $3x-2y+5=0$
• D. $3x-2y+10=0$

Answer 1

The correct option is C $3x-2y+5=0$


$m_{P{P}^{\prime }}=\frac{4}{-6}=\frac{-2}{3}\Rightarrow m_L=\frac{3}{2}$

Required line $L$ is the perpendicular bisector of $P{P}^{\prime }$
So, equation of $L=0$ is
$y-4=\frac{3}{2}(x-1)\Rightarrow 3x-2y+5=0$