Question

The perpendicular bisector of the line segment connecting the points (-3,8) and (-5,4) has an equation of the form y = mx + b. Find m+b. BTW, the answer is not 16...

Answer 1

Explanation:

find the slope

`(4-8)/(-5-(-3)) =(-4)/(-2) \\\\slope=2\\y=mx+b\\y=2x+b\\`

take a coordinate to fill in

`(-5,4)\\y=-5\\x=4\\-5=2(4)+b\\-5=8+b-8 -8\\-13=b\\`

this means that the equation is y=2x-13

and if you add m and b

you get :-11

I HOPE THIS HELPS

Answer 2

7/2

Explanation:

Let $A = (-3,8)$ and $B = (-5,4)$. The midpoint of $\overline{AB}$ is $\left( \frac{(-3) + (-5)}{2}, \frac{8 + 4}{2} \right) = (-4,6)$.

The slope of $\overline{AB}$ is $\frac{8 - 4}{(-3) - (-5)} = 2$, so the slope of the perpendicular bisector of $\overline{AB}$ is $-\frac{1}{2}$. Therefore, the equation of the perpendicular bisector is given by

\[y - 6 = -\frac{1}{2} (x + 4).\]Isolating $y,$ we find

\[y = -\frac{1}{2} x + 4.\]