Please Help me with this Math question A.S.A.P.?

Question

Triangle PQR has vertices P(0,4), Q(8,-2), R(7,5). Write equations for the perpendicular bisectors of the 3 sides of the triangle. And show the perpendicular bisectors intersect at a single point C, called the circumcenter of the triangle.

Answer 1

Greetings,

PQ slope = -6/8 = - 3/4

slope of perpendicular = 4/3

midpoint of PQ = (4,1)

equation of perpendicular bisector of PQ

y - 1 = (4/3)(x - 4)

3y - 3 = 4x - 16

4x - 3y = 13

PR slope = 1/7

slope of perpendicular = -7

midpoint of PR = (7/2,9/2)

equation of perpendicular bisector of PR

y - 9/2 = -7(x - 7/2)

2y - 9 = -14x + 49

14x + 2y = 58

7x + y = 29

QR slope = -7

slope of perpendicular = 1/7

midpoint of QR = (15/2,3/2)

equation of perpendicular bisector of QR

y - 3/2 = (1/7)(x - 15/2)

14y - 21 = 2x - 15

2x - 14y = -6

x - 7y = -3

These intersect at the point (4,1), also since the slopes of PR and QR are reciprocals, this is a right angled triangle and the circumcenter lies on the midpoint of the hypoteneuse PQ

Regards