Question

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

Answer 1

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

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

$Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)$

$Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)$

$Midpoint=\left(-4,1\right)$

Slope of lines NY is

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

$m=\dfrac{-3-5}{3-(-11)}$

$m=\dfrac{-8}{14}$

$m=\dfrac{-4}{7}$

Product of slopes of two perpendicular lines is -1. So,

$m_1\times \dfrac{-4}{7}=-1$

$m_1=\dfrac{7}{4}$

The perpendicular bisector of NY passes through (-4,1) with slope $\dfrac{7}{4}$. So, the equation of perpendicular bisector of NY is

$y-y_1=m_1(x-x_1)$

$y-1=\dfrac{7}{4}(x-(-4))$

$y-1=\dfrac{7}{4}(x+4)$

$y-1=\dfrac{7}{4}x+7$

Add 1 on both sides.

$y=\dfrac{7}{4}x+8$

Therefore, the equation of perpendicular bisector of NY is $y=\dfrac{7}{4}x+8$.