Question

find an equation for the perpendicular bisector of the line segment whose endpoints are $(-9,-3)$ and $(5,1)$.

Explanation:

Step1: Find the mid - point of the line segment

The mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
For the points \((-9,-3)\) and \((5,1)\), we have \(x_1=-9,x_2 = 5,y_1=-3,y_2 = 1\).
\(x=\frac{-9 + 5}{2}=\frac{-4}{2}=-2\), \(y=\frac{-3+1}{2}=\frac{-2}{2}=-1\).
The mid - point is \((-2,-1)\).

Step2: Find the slope of the line segment

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
\(m=\frac{1-(-3)}{5-(-9)}=\frac{1 + 3}{5 + 9}=\frac{4}{14}=\frac{2}{7}\).

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 \(m_1\).
Since \(m\times m_1=-1\) and \(m = \frac{2}{7}\), then \(m_1=-\frac{7}{2}\).

Step4: Use the point - slope form to find the equation of the perpendicular bisector

The point - slope form is \(y - y_0=m_1(x - x_0)\), where \((x_0,y_0)=(-2,-1)\) and \(m_1=-\frac{7}{2}\).
\(y+1=-\frac{7}{2}(x + 2)\).
Expand: \(y+1=-\frac{7}{2}x-7\).
\(y=-\frac{7}{2}x-8\).

Answer:

\(y =-\frac{7}{2}x - 8\)