Question

Find the equation of the perpendicular bisector of the line joining $(1,6)$ and $(2,-3)$

Answer 1

Let the point be $A(1,6)$ and $B(2,-3)$

Midpoint of $AB=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{-3+6}{2},\frac{2+1}{2})$

$=(\frac{3}{2},\frac{3}{2})$

Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-6}{2-1}=-9$

$\therefore$Slope of the perpendicular$=\frac{1}{9}$

Now equation is $y-y_1=m(x-x_1)$

$y-\frac{3}{2}=\frac{1}{9}(x-\frac{3}{2})$

$\Rightarrow 9y-\frac{27}{2}=x-\frac{3}{2}$

$\Rightarrow x-9y-\frac{3}{2}+\frac{27}{2}=0$

$\Rightarrow x-9y+\frac{24}{2}=0$

$\Rightarrow x-9y+12=0$

$\therefore x-9y+12=0$ is the equation of the line.