Question

Find an equation for the perpendicular bisector of the line segment whose endpoints(-4,3) and (8,9).are

Answer 1

Given:

Endpoints of the line segment

(-4,3) and (8,9).

Required:

equation for the perpendicular bisector of the line segment

Solution:

First, we have to know the equation of the line segment using the two-point form:

`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`

P1 (-4,3) and P2 (8,9).

`\begin{gathered} y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1) \\ y-3=(9-3)/(8-(-4))\lbrack x-(-4)\rbrack \\ y-3=(6)/(12)(x+4) \\ y-3=(1)/(2)(x+4) \\ 2\cdot\lbrack y-3=(1)/(2)(x+4)\rbrack\cdot2 \\ 2y-6=x+4 \\ _{}x-2y+4+6=0 \\ x-2y+10=0 \end{gathered}`

Then, we calculate for the midpoint of the line segment. The formula for the midpoint M( xm,ym ) is:

`(x_m,y_m)=((x_1+x_2)/(2),(y_1+y_2)/(2))`

Using P1 (-4,3) and P2 (8,9), the coordintes of the midpoint are

`\begin{gathered} (x_m,y_m)=((x_1+x_2)/(2),(y_1+y_2)/(2)) \\ (x_m,y_m)=((-4+8)/(2),\frac{3+9_{}}{2}) \\ (x_m,y_m)=((4)/(2),\frac{12_{}}{2}) \\ (x_m,y_m)=(2,6) \end{gathered}`

The bisector divides the line segment into two equal parts ( it intersects the line at the midpoint. It is perpendicular to the line segment.

We know that perpendicular lines have opposite-reciprocal slopes.

The slope of the line segment is :

`_{}(y_2-y_1)/(x_2-x_1)=(9-3)/(8-(-4))=(6)/(12)=(1)/(2)`

Thus, the slope of the perpendicular bisector is the negative reciprocal of 1/2

`m=-(1)/((1)/(2))=-2`

At this point we can now determine the equation of the bisector using the midpoint and the slope of the perpendicular bisector

`\begin{gathered} M(2,6) \\ m=-2 \end{gathered}`

The point-slope form a line is:

`\begin{gathered} y-y_1=m(x-x_1) \\ y-6=-2(x-2) \\ y-6=-2x+4 \\ 2x+y-10=0 \end{gathered}`

Answer:

The equation of the perpendicular bisector is 2x + y - 10 = 0