Question

Find the equation of the perpendicular bisector of the line joining $(1,3)$ and $(-2,4)$

Answer 1

Let the point be $A(1,3)$ and $B(-2,4)$

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

$=(\frac{-2+1}{2},\frac{3+4}{2})$

$=(\frac{1}{2},\frac{7}{2})=(\frac{1}{2},\frac{7}{2})$

Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{3-4}{-2-1}=\frac{-1}{-3}=\frac{1}{3}$

$\therefore$Slope of the perpendicular$=-3$

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

$y-\frac{7}{2}=-3(x-\frac{1}{2})$

$\Rightarrow y-\frac{7}{2}=-3x+\frac{3}{2}$

$\Rightarrow y-\frac{7+3}{2}+3x=0$

$\Rightarrow 3x+y-\frac{10}{2}=0$

$\Rightarrow 3x+y-5=0$

$\therefore 3x+y-5=0$ is the required equation of the line.