Determine the point of intersection of right bisectors in a triangle ∆𝐴𝐵𝐶 with vertices A (-3, 5), B (1, 1) and 𝐶(−7, −3). Find the distance from the point of intersection to each vertex of the triangle.
+128474 Find the midpoint of AB = (-1, 3)
Find the slope of AB = 4/ -4 = -1
A perpendicular line to AB through the midpoint of AB has the equation
y = 1 (x - -1) + 3
y = x + 4 (1)
Find the midpoint of BC = ( -3, -1)
Slope of BC = ( -4/ -8) = 1/2
A perpedicular line to BC through the midpoint of BC has the equation
y = (2) ( x - - 3) -1
y = (-2)( x + 3) - 1
y = -2x -7 (2)
Find the x intersection of (1) and (2)
x + 4 = -2x - 7
3x = -11
x = -11/3
And y = (-11/3) + 4 = 1/3
So.....the point of intersection ( the circumcenter) = ( -11/3, 1/3)
The distance from this point to each vertex is the same =
sqrt [ ( -11/3 - 1)^2 + (1/3 - 1)^2 ] = sqrt [ (14/3)^2 + ( 2/3)^2 ] = sqrt [ 14^2 + 2^2] / 3 =
sqrt [ 200] / 3 = 10sqrt (2) / 3
