Original post by nerd434Now for b), I'm not certain as to whether this solution is suffice. Is there another way of doing it? If so, please explain.
Question: Triangle ABC vertices (-2,5) , (6,1) and (2,-7) respectively.
a) Find the equation of the perpendicular bisector of BC.
b) Show that this perpendicular bisector passes through the mid-point of side AC.
Solution:
a) Midpoint BC = (6+2/2) , (1-7/2) = (4,-3)
mBC = (-7 - 1) / (2 - 6) = -8/-4 = 2.
Thus mI = -1/2
Equation:
y - b = m (x-a)
y - (-3) = -1/2 (x-4)
2(y+3) = -1 (x-4)
2y + 6 = -1x +4
2y = -x -2
Perpendicular Bisector Equation: y = (-1/2 x) - 1
b) Midpoint AC = (-2+2/2) , (5-7/2) = (0, -1)
If the perpendicular bisector passes through the midpoint of AC, when the value of x is 0, y = -1 and when the value of y = -1, the value of x = 0, y = -1.
y = (-1/2)x - 1
y = (-1/2)(0) - 1
y = -1
y = (-1/2)x - 1
-1 = (-1/2)x - 1
0 = (-1/2)x
x = 0.
As the equation of the perpendicular bisector satisfies the midpoint of AC, the lines must cut through the point (0, -1).