Question

find an equation for the perpendicular bisector of the line segment whose endpoints are (-2,2) and (-8,8).

Explanation:

Step1: Find the mid - point of the line segment

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.
For $(x_1=-2,y_1 = 2)$ and $(x_2=-8,y_2 = 8)$, the mid - point $M$ is $(\frac{-2+( - 8)}{2},\frac{2 + 8}{2})=(\frac{-10}{2},\frac{10}{2})=(-5,5)$.

Step2: Find the slope of the line segment

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$.
For $(x_1=-2,y_1 = 2)$ and $(x_2=-8,y_2 = 8)$, the slope $m_1=\frac{8 - 2}{-8-( - 2)}=\frac{6}{-6}=-1$.

Step3: Find the slope of the perpendicular bisector

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the perpendicular bisector be $m_2$. Since $m_1=-1$ and $m_1m_2=-1$, then $m_2 = 1$.

Step4: Find the equation of the perpendicular bisector

Use the point - slope form of a line $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is a point on the line (the mid - point $(-5,5)$) and $m$ is the slope of the line ($m = 1$).
$y-5=1\times(x + 5)$
$y-5=x + 5$
$y=x+10$

Answer:

$y=x + 10$