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:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

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

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

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

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

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

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

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 :

$_{}\frac{y_2-y_1}{x_2-x_1}=\frac{9-3}{8-(-4)}=\frac{6}{12}=\frac{1}{2}$

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

$m=-\frac{1}{\frac{1}{2}}=-2$

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