The equation in slope intercept form of a line that is a perpendicular bisector of segment AB with endpoints A(-5,5) and B(3,-3) is y = x + 2
Solution:
Given, two points are A(-5, 5) and B(3, -3)
We have to find the perpendicular bisector of segment AB.
Now, we know that perpendicular bisector passes through the midpoint of segment.
The formula for midpoint is:
[tex]\text { midpoint }=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)[/tex]
[tex]Here x_1 = -5 ; y_1 = 5 ; x_2 = 3 ; y_2 = -3[/tex]
[tex]\text { So, midpoint of } A B=\left(\frac{-5+3}{2}, \frac{5+(-3)}{2}\right)=\left(\frac{-2}{2}, \frac{2}{2}\right)=(-1,1)[/tex]
Finding slope of AB:
[tex]\text { Slope of } A B=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]\text { Slope } m=\frac{-3-5}{3-(-5)}=\frac{-8}{8}=-1[/tex]
We know that product of slopes of perpendicular lines = -1
So, slope of AB [tex]\times[/tex] slope of perpendicular bisector = -1
- 1 [tex]\times[/tex] slope of perpendicular bisector = -1
Slope of perpendicular bisector = 1
We know its slope is 1 and it goes through the midpoint (-1, 1)
The slope intercept form is given as:
y = mx + c
where "m" is the slope of the line and "c" is the y-intercept
Plug in "m" = 1
y = x + c ---- eqn 1
We can use the coordinates of the midpoint (-1, 1) in this equation to solve for "c" in eqn 1
1 = -1 + c
c = 2
Now substitute c = 2 in eqn 1
y = x + 2
Thus y = x + 2 is the required equation in slope intercept form
