Question

What's the equation of the line that's a perpendicular bisector of the segment connecting A (–2, 8) and B (–4, 2)?

Question 12 options:

A)

y = –1∕3x – 3

B)

y = –1∕3x + 3

C)

y = 1∕3x + 3

D)

y = –1∕3x + 4

Answer 1

Explanation:

D)

y = –1∕3x + 4

Answer 2

D

Explanation:

The required line is perpendicular bi sector. So it goes through the midpoint of AB. Find the midpoint of AB

Midpoint
`((x_(1)+x_(2))/(2) , (y_(1)+y_(2))/(2))\\`

Midpoint of AB =
`((-2 + [-4])/(2) , (8+2)/(2))`

`=((-6)/(2) , (10)/(2))\\\\=(-3 , 5)`

Find the slope of AB,
`m_(1)`

slope =
`(y_(2)-y_(1))/(x_(2)-x_(1))\\\\`

`=(2 - 8)/(-4-[-2])\\\\=(-6)/(-4+2)\\\\=(-6)/(-2)\\\\=3`

Slope of the perpendicular line to AB =
`(-1)/(m_(1))` = -1/3

The required line passes through (-3, 5) and has slope -1/3

Equation of required line : y - y1 = m (x -x1)

`y - 5 = (-1)/(3) (x - [-3])\\\\y-5=(-1)/(3) ( x +3)\\\\y-5=(-1)/(3)x + 3 *(-1)/(3)\\\\y-5=(-1)/(3)x-1\\\\y=(-1)/(3)x-1+5\\\\y=(-1)/(3)x + 4`