Question

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

Answer 1

`y = (1)/(2) x + 3`

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 = (1)/(2) x + 3`

Answer 2

First find the equation of y.

`y=mx+n=(\Delta y)/(\Delta x)x+n`

Find the slope m.

`m=(5-9)/(-1-3)=(-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=-(1)/(m)=-1`.

The equation of a perpendicular bisector is thus:

`y_(\perp)=-x+12`.

Hope this helps.