An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,5 )# to #(8 ,2 )# and the triangle's area is #36 #, what are the possible coordinates of the triangle's third corner?

Refer diagram.

#a = sqrt((8-7)^2 + (2-5)^2) = 3.1623#

Midpoint of base BC = #(8+7)/2, (2+5)/2 = (15/2, 7/2)#

Slope of BC #m_(BC) = (2-5) / (8-7) = -3#

Slope of altitude AD of the triangle passing through the midpoint D is #m_(AD) = -1/(m_(BC)) = -(1/-3) = 1/3#

Equation of altitude AD is

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

#6y - 21 = 2x - 15#

#3y - x = 3#. #color(blue)(Eqn) color(white)(xx)color (blue)(1)#

Area of triangle #A = (1 / 2) a h#

#h = (2 * 36) / 3.1623 = 22.7682#

Slope of AB = #tan theta = m_(AB) = h / (a/2) = 22.7682 / (3.1623/2) = 14.4#

Equation of AB is

#y - 5 = 14.4 * (x - 7)#

#y - 14.4x = -95.8#. #color(blue)(Eqn) color(white)(xx)color (blue)(2)#

Solving Eqns (1) & (2) we get the coordinates of vertex ‘A’

#color(purple)( A (1452/211), (695/211))#

Verification :

Length of #AB = b = sqrt((7 - (1452/211))^2 + (5 - (695/211)^2) = 1.7103#

Length of #AC = b = sqrt((8-(1452/211))^2 + (2 - (695/211)’^2) = 1.7103#

Value of side #AB = AC = 1.7103#.

Hence its an isosceles triangle with sides #color(green)(3.1623, 1.7103, 1.7103)#