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

Answer:

D

Step-by-step explanation:

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

Midpoint $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$

Midpoint of AB = $(\frac{-2 + [-4]}{2} , \frac{8+2}{2})$

                        $=(\frac{-6}{2} , \frac{10}{2})\\\\=(-3 , 5)$

Find the slope of AB, $m_{1}$

slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

         $=\frac{2 - 8}{-4-[-2]}\\\\=\frac{-6}{-4+2}\\\\=\frac{-6}{-2}\\\\=3$

Slope of the perpendicular line to AB = $\frac{-1}{m_{1}}$   = -1/3

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

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

$y - 5 = \frac{-1}{3} (x - [-3])\\\\y-5=\frac{-1}{3} ( x +3)\\\\y-5=\frac{-1}{3}x + 3 *\frac{-1}{3}\\\\y-5=\frac{-1}{3}x-1\\\\y=\frac{-1}{3}x-1+5\\\\y=\frac{-1}{3}x + 4$

Answer 2

Answer:

Step-by-step explanation:

D)

y = –1∕3x + 4