Question

What's the equation of the line that's a perpendicular bisector of the segment

connecting A(-2, 8) and B(-4, 2)?

Answer 1

y = -
`(1)/(3)` x + 4

Explanation:

The perpendicular bisector bisects AB at right angles

The midpoint of AB using the midpoint formula

`M_(AB)` = (
`\frac{x_{1+x_(2) } }{2}` ,
`(y_(1)+y_(2) )/(2)` )

with (x₁,y₁ ) = A (- 2, 8 ) and (x₂, y₂ ) = B (- 4, 2 )

`M_(AB)` = (
`(-2-4)/(2)` ,
`(8+2)/(2)` ) = (
`(-6)/(2)` ,
`(10)/(2)` ) = (- 3, 5 )

Calculate the slope m of AB using the slope formula

m =
`(y_(2)-y_(1) )/(x_(2)-x_(1) )` =
`(2-8)/(-4-(-2))` =
`(-6)/(-4+2)` =
`(-6)/(-2)` = 3

Given a line with slope m then the slope of a line perpendicular to it is

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/(3)`

The perpendicular bisector passes through (- 3, 5 ) with slope = -
`(1)/(3)`

The equation in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept ) , then

y = -
`(1)/(3)` x + c ← is the partial equation

To find c substitute (- 3, 5 ) into the partial equation

5 = 1 + c ⇒ c = 5 - 1 = 4

y = -
`(1)/(3)` x + 4 ← equation of perpendicular bisector