Question

find an equation for the perpendicular bisector of the line segment whose endpoints are (5, 2) and (-3, 6).

Explanation:

Step1: Find the mid - point

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 $(x_1 = 5,y_1 = 2)$ and $(x_2=-3,y_2 = 6)$, the mid - point $M$ is $(\frac{5+( - 3)}{2},\frac{2 + 6}{2})=(1,4)$.

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 $(x_1 = 5,y_1 = 2)$ and $(x_2=-3,y_2 = 6)$, the slope $m_1=\frac{6 - 2}{-3 - 5}=\frac{4}{-8}=-\frac{1}{2}$.

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 $m_1\times m_2=-1$ and $m_1 = -\frac{1}{2}$, then $-\frac{1}{2}\times m_2=-1$, so $m_2 = 2$.

Step4: Find the equation of the perpendicular bisector

The point - slope form of a line is $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope.
Using the point $(x_0 = 1,y_0 = 4)$ and $m = 2$, we have $y - 4=2(x - 1)$.
Expand to get $y-4=2x-2$, or $y = 2x+2$.

Answer:

$y = 2x + 2$