Midpoint of AB = ( (5 − 2) / 2 , ( +2) /2 ) = ( 3/2 , 5/2) = ( 1.5, 2.5)
Slope of AB = [ 3 − 2] / [ 5 − − 2] = [1 / 7] .... − reciprocal slope..... = −7
Equation of perp. bisec. = y = − 7 (x − 1.5) + 2.5 → y = − 7x + 10.5 + 2.5 →
y = − 7x + 13 (1)
Midpoint of BC = ( (2 − 2) / 2 , (−6 +2) /2 ) = ( 0/2 , −4/2) = ( 0, −2)
Slope of BC = [ 2 − − 6] / [ −2 − 2] = [8 / −4] = −2 .... − reciprocal slope..... = 1/2
Equation of perp. bisec. = y = (1/2) (x − 0) − 2 → y = (1/2)x − 2 (2)
Midpoint of AC = ( (5 + 2) / 2 , (−6 +3) /2 ) = ( 7/2 , −3/2) = ( 3.5, −1.5)
Slope of AC = [ 3 − − 6] / [ 5 − 2] = [9 / 3] = 3 .... − reciprocal slope..... = −1/3
Equation of perp. bisec. = y = (−1/3) (x − 3.5) − 1.5 → y = (−1/3)x + 7/6 − 1.5 →
y = (−1/3)x − 1/3 (3)
Intersection of (1) and (2)
− 7x + 13 = (1/2)x − 2 add 2, 7x to both sides.....
15 = (1/2)x + 7x
15 = 7.5x divide both sides by 7.5
2 = x and y = − 7(2) + 13 = −1 = ( 2, −1)
And y = (−1/3)x − 1/3 also passes through the point ( 2, −1) → y = (−1/3)(2) − 1/3 =
−2/3 − 1/3 = −1
Distance from ( 2, −1) to A = √[ ( 2−5)^2 + ( −1 − 3)^2 ] = 5
Distance from ( 2, −1) to B = √[ ( 2− − 2)^2 + ( −1 − 2)^2 ] = 5
Distance from ( 2, −1) to C = √[ ( 2− 2)^2 + ( −1 − − 6)^2 ] = 5
I'll leave it to you, Tony, to check the distance computations above
