A triangle has corners at #(3,7)#, #(4,1)#, and #(8,2)#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

A, B, C are the vertices and Let D, E, F be the mid points of the sides a, b, c respectively.

Midpoint of BC = D = (x2 + x1)/2, (y2 + y1)/2 = (4+8)/2, (1+2)/2 =
(6,3/2)#

Slope of BC #m_(BC) = (y2-y1)/(x2-x1) = (2-1)/(8-4) = 1/4#

Slope of the perpendicular bisector through D = #-1/m_(BC) = -4#

Equation of perpendicular bisector passing through mid point D using standard form of equation #y - y1 = m * (x - x1)#

#y-(3/2) = -4 * (x - 6)#

#color(red)(2y + 8x = 51)# Eqn (D)

Similarly,

Mid point of CA = #E (11/2, 9/2)#

Slope of CA = m_(CA) = (2-7)/(8-3) = -1#

Slope of the perpendicular bisector through E = #-1/m_(CA) = 1#

Equation of perpendicular bisector passing through mid point E is

#y-(9/2) = 1 * (x - (11/2)#

#color(red)(y - x = -1)# Eqn (E)

Similarly,

Mid point of AB = #F (7/2, 4)#

Slope of AB= m_(AB) = (1-7)/(4-3) = -6#

Slope of the perpendicular bisector through F = #-1/m_(AB) = 1/6#

Equation of perpendicular bisector passing through mid point F is

#y - 4 = (1/6) * (x - (7/2)#

#12y - 48 = 2x - 7#

#color(red)(12y - 2x = 41)# Eqn (F)

Solving Eqns (D), (E), we get the coordinates of circumcenter P.

#color(green)(P (53/10, 43/10))#

This can be verified by solving Eqns (E), (F).
and the answer is #color(green)(P (53/10, 43/10)#

Length of the perpendicular bisectors PD

#PD = sqrt((6-(53/10))^2 + ((3/2) - (43/10))^2) = color(blue)(2.8862)#

Length of perpendicular bisector PE

#EP = sqrt(((11/2) - (53/10))^2 + ((9/2) - (43/10))^2) = color(blue)(0.2828)#

Length of perpendicular bisector PF
7/2, 4
#PF = sqrt(((7/2) - (53/10))^2 + (4 - (43/10))^2) = color(blue)(1.8248)#