Given in isosceles #\triangle ABC#, the side #A# opposite to the vertex #A# goes through the vertices #B(4, 1)# & #C(8, 5)#.
#BC=\sqrt{(4-8)^2+(1-5)^2}#
#=4\sqrt2#
Mid-point say #M# of side #BC:\ \ \ M(\frac{4+8}{2}, \frac{1+5}{2})\equiv(6, 3)#
Let #h# be the length of altitude drawn from vertex #A# to the side #BC# then area of #\triangle ABC#
#1/2(BC)h=64#
#1/2(4\sqrt2)h=64#
#h=16\sqrt2#
Let the third vertex be #A(x, y)# then the altitude #AM# drawn from vertex #A(x, y)# to the midpoint #M(6, 3)# of side #BC# will be perpendicular to the side #BC# hence the condition of perpendicularity of two lines
#\frac{y-3}{x-6}\times \frac{5-1}{8-4}=-1#
#y=-x+9\ \ ...........(1)#
Now, the distance between the vertex #A(x, y)# & mid-points #M(6, 3)# of side BC must be equal to the length of altitude #h=16\sqrt2# drawn from vertex A to the side BC. Now, using distance formula
#\sqrt{(x-6)^2+(y-3)^2}=16\sqrt2#
#(x-6)^2+(y-3)^2=512#
#(x-6)^2+(-x+9-3)^2=512\ \quad (\because y=-x+9\ \text{from (1)})#
#(x-6)^2=256#
#x-6=\pm16#
#x=6\pm 16#
#x=22, -10#
setting these values of #x# into (1), we get corresponding values of #y# as follows
#y=-22+9=-13 # &
#y=-(-10)+9=19 #
Hence the coordinates of third vertex are #A(22, -13)# or #A(-10, 19)#
