Question

Nov 30 5:29:05 PMWatch help videoFind an equation for the perpendicular bisector of the line segment whose endpointsare(-8 -3) and (2,3).Sihmtes

Answer 1

The form of the equation of the line is

y = m x + b, where

m is the slope

b is the y-intercept

The rule of the slope is

`m=(y2-y1)/(x2-x1)`

(x1, y1) and (x2, y2) are two points on the line

To find the equation of the perpendicular bisector to the line whose endpoints are (-8, -3) and (2, 3), we must find the slope of this line

x1 = -8 and y1 = -3

x2 = 2 and y2 = 3

Substitute them in the rule above

`m=(3--3)/(2--8)=(3+3)/(2+8)=(6)/(10)=(3)/(5)`

To find the slope of the perpendicular line reciprocal the fraction and change its sign

The slope of the perpendicular line is

`-(5)/(3)`

Substitute it in the form of the equation

`y=-(5)/(3)x+b`

To find b we must have a point on the line

Since the line is the bisector of the given line, then it passes through its mid-point, then we need to find the mid-point of the given line

`M=((x1+x2)/(2),(y1+y2)/(2))`

We will use the endpoints above

`\begin{gathered} M=((-8+2)/(2),(-3+3)/(2)) \\ M=(-(6)/(2),(0)/(2)) \\ M=(-3,0) \end{gathered}`

The mid-point is (-3, 0)

We will use it to find b

Substitute x by -3 and y by 0 in the equation to find b

`\begin{gathered} 0=-(5)/(3)(-3)+b \\ 0=5+b \end{gathered}`

Subtract 5 from both sides

-5 = b + 5 - 5

-5 = b

Substitute it in the equation

`\begin{gathered} y=-(5)/(3)x+(-5) \\ y=-(5)/(3)x-5 \end{gathered}`

The equation of the perpendicular bisector is

`y=-(5)/(3)x-5`