Question

Find the equation of the perpendicular bisector of AB when A(-7,2) and B(3,4)

Answer 1

Given:

Two points are A(-7,2) and B(3,4).

To find:

The perpendicular bisector of AB.

Solution:

Slope formula:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Slope of AB is

$m_1=\dfrac{4-2}{3-(-7)}$

$m_1=\dfrac{2}{3+7}$

$m_1=\dfrac{2}{10}$

$m_1=\dfrac{1}{5}$

Procut of slopes of perpendicular line is -1.

So, slope of perpendicular bisect is opposite of reciprocal of $\dfrac{1}{5}$.

$m_2=-5$

Midpoint of AB is

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-7+3}{2},\dfrac{2+4}{2}\right)$

$Midpoint=\left(\dfrac{-4}{2},\dfrac{6}{2}\right)$

$Midpoint=\left(-2,3\right)$

Slope of perpendicular bisector is -5 and it passes through (-2,3), so the equation of perpendicular bisector is

$y-y_1=m(x-x_1)$

where, m is slope.

$y-3=-5(x-(-2))$

$y-3=-5(x+2)$

$y-3=-5x-10$

Add 3 on both sides.

$y=-5x-10+3$

$y=-5x-7$

Therefore, the equation of perpendicular bisector is $y=-5x-7$.