Question

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

Answer 1

Answer:

$y=-3x-8$

Step-by-step explanation:

Hi there!

What we need to know:

• Midpoint: $(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )$ where the endpoints are $(x_1,y_1)$ and $(x_2,y_2)$

• Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
• Perpendicular lines always have slopes that are negative reciprocals (ex. 2 and -1/2, 3/4 and -4/3, etc.)

1) Determine the midpoint of the line segment

When two lines bisect each other, they intersect at the middle of each line, or the midpoint.

$(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )$

Plug in the endpoints (5, -3) and (-7, -7)

$(\frac{5+(-7)}{2} ,\frac{-3+(-7)}{2} )\\(\frac{-2}{2} ,\frac{-10}{2} )\\(-1,-5)$

Therefore, the midpoint of the line segment is (-1,-5).

2) Determine the slope of the line segment

Recall that the slopes of perpendicular lines are negative reciprocals. Doing this will help us determine the slope of the perpendicular bisector.

Slope = $\frac{y_2-y_1}{x_2-x_1}$ where the given points are $(x_1,y_1)$ and $(x_2,y_2)$

Plug in the endpoints (5, -3) and (-7, -7)

$\frac{-7-(-3)}{-7-5}\\\frac{-7+3}{-7-5}\\\frac{-4}{-12}\\\frac{1}{3}$

Therefore, the slope of the line segment is $\frac{1}{3}$. The negative reciprocal of $\frac{1}{3}$ is -3, so the slope of the perpendicular is -3. Plug this into $y=mx+b$:

$y=-3x+b$

3) Determine the y-intercept of the perpendicular bisector (b)

$y=-3x+b$

Recall that the midpoint of the line segment is is (-1,-5), and that the perpendicular bisector passes through this point. Plug this point into $y=-3x+b$ and solve for b:

$-5=-3(-1)+b\\-5=3+b$

Subtract 3 from both sides

$-5-3=3+b-3\\-8=b$

Therefore, the y-intercept of the line is -8. Plug this back into $y=-3x+b$:

$y=-3x-8$

I hope this helps!