Triangle ABC has the following points: A (-2,-2), B (4,4), C (18,-6). Use these points to write the equations of the line containing the perpendicular bisector of AC¯¯¯¯¯¯¯¯ in point slope form. Make sure to show all work in order to receive full points.
To find the equation of the line containing the perpendicular bisector of AC, we need to first find the midpoint of AC.
The midpoint can be found by finding the average of the x-coordinates and the average of the y-coordinates of A and C.
Midpoint of AC = ( (x-coordinate of A + x-coordinate of C)/2 , (y-coordinate of A + y-coordinate of C)/2 )
Midpoint of AC = ( (-2 + 18)/2 , (-2 + -6)/2 )
Midpoint of AC = ( 8 , -4 )
Now, let's find the slope of AC.
Slope of AC = (y-coordinate of C - y-coordinate of A) / (x-coordinate of C - x-coordinate of A)
Slope of AC = (-6 - (-2)) / (18 - (-2))
Slope of AC = (-6 + 2)/(18 + 2)
Slope of AC = -4/20
Slope of AC = -1/5
Since the perpendicular bisector of AC is perpendicular to AC, its slope will be the negative reciprocal of the slope of AC. In other words, the slope of the perpendicular bisector of AC is 5.
Now, we have the slope (m = 5) and the midpoint (8,-4). We can use the point-slope form of a line to write the equation of the perpendicular bisector of AC.
Point-slope form: y - y1 = m(x - x1)
Substituting the values:
y - (-4) = 5(x - 8)
Simplifying:
y + 4 = 5x - 40
Rewriting in point-slope form:
y = 5x - 44
Therefore, the equation of the line containing the perpendicular bisector of AC in point-slope form is y = 5x - 44.