Question

Find the equation of the perpendicular bisector of a line segment whose end points are (-3,9) and (-1,5)

Answer 1

First find the equation of y.

$y=mx+n=\frac{\Delta y}{\Delta x}x+n$

Find the slope m.

$m=\frac{5-9}{-1-3}=\frac{-4}{-4}=1$.

Pick one point, I'll pick (-3, 9). Insert coordinates in equation then compute n.

$9=1(-3)+n\implies n=12$.

The equation of a line y is:

$y=x+12$.

The perpendicular line $y_{\perp}$ is same like the normal line except its slope m becomes:

$k=-\frac{1}{m}=-1$.

The equation of a perpendicular bisector is thus:

$y_{\perp}=-x+12$.

Hope this helps.

Answer 2

Answer:

$y = \frac{1}{2} x + 3$

Step-by-step explanation:

Using the given points, first find the slope:

m = (₂ - y₁) / (x₂ - x₁)

   = (5 - 9) / (-1 - (-3))

   = -4 / 2

    = -2

Find y-intercept by using anyone of the give points and slope from above:

y = mx + b

5 = -2(-1) + b

5 = 2 + b

b = 3

With the m and b, you can make equation of line:

y = mx + b

y = -2x + 3

To find the perpendicular slope, make it opposite and inverse it.

m = -2 becomes 1/2

Regular equation of line:

y = -2x + 3

Perpendicular equation of line:

$y = \frac{1}{2} x + 3$