Question

Find an equation for the perpendicular bisector of the line segment whose endpoints

Answer 1

Answer:

The equation for the perpendicular bisector of the line segment will be:

$y=-\frac{7}{2}x-13$

Step-by-step explanation:

Given the endpoints of the line segments

• (5,-4)

• (-9, -8)

Determining the slope between (5,-4) and (-9, -8)

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(5,\:-4\right),\:\left(x_2,\:y_2\right)=\left(-9,\:-8\right)$

$m=\frac{-8-\left(-4\right)}{-9-5}$

$m=\frac{2}{7}$

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:  

slope = m = 2/7

Thus, the slope of the the new perpendicular line = – 1/m = (-1)/(2/7)= -7/2

Next, determining the mid-point between  (5,-4) and (-9, -8)

$\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$

$\left(x_1,\:y_1\right)=\left(5,\:-4\right),\:\left(x_2,\:y_2\right)=\left(-9,\:-8\right)$

$=\left(\frac{-9+5}{2},\:\frac{-8-4}{2}\right)$

Refine

$=\left(-2,\:-6\right)$

We know that the point-slope form of equation of line is

$y-y_1=m\left(x-x_1\right)$

where

• m is the slope of the line

• (x₁, y₁) is the point

substituting the slope of the perpendicular line -7/2 and the point (-2, -6)

$y-y_1=m\left(x-x_1\right)$

$y-\left(-6\right)=-\frac{7}{2}\left(x-\left(-2\right)\right)$

$y+6=-\frac{7}{2}\left(x+2\right)$

Subtract 6 from both sides

$y+6-6=-\frac{7}{2}\left(x+2\right)-6$

$y=-\frac{7}{2}x-13$

Therefore, the equation for the perpendicular bisector of the line segment will be:

$y=-\frac{7}{2}x-13$