Question

find an equation for the perpendicular bisector of the line segment whose endpoints are (-1, -1) and (9, -5).

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 points $(-1,-1)$ and $(9,-5)$, the mid - point $M$ is $(\frac{-1 + 9}{2},\frac{-1+( - 5)}{2})=(4,-3)$.

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 points $(-1,-1)$ and $(9,-5)$, the slope $m_1=\frac{-5-( - 1)}{9-( - 1)}=\frac{-4}{10}=-\frac{2}{5}$.

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 $m_2$. Then $m_1\times m_2=-1$.
Since $m_1 = -\frac{2}{5}$, we have $-\frac{2}{5}\times m_2=-1$, so $m_2=\frac{5}{2}$.

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 the mid - point $(4,-3)$ and $m=\frac{5}{2}$.
$y-( - 3)=\frac{5}{2}(x - 4)$
$y + 3=\frac{5}{2}x-10$
$y=\frac{5}{2}x-13$

Answer:

$y=\frac{5}{2}x - 13$