An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(4 ,9 )# to #(8 ,5 )# and the triangle's area is #64 #, what are the possible coordinates of the triangle's third corner?

1 Answer
May 5, 2016

The third corner will be at either #(-2,-1)# or #(14,15)#

Explanation:

Length of side A = #sqrt((8-4)^2+(5-9)^2) = 4sqrt(2)#

Height of triangle ABC relative to A #= ("area")/("length side A")#
#color(white)("XXX")64/(4sqrt(2))= 8sqrt(2)#

Note that since #abs(B)=abs(C)# the height relative to #A# is the length of the perpendicular bisector of #A#

Mid-point of #A# is #((4+8)/2,(9+5)/2)=(6,7)#
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#A# has a slope of #(Deltay)/(Deltax)=(9-5)/(4-8)=-1#

#rArr# the perpendicular bisector has a slope of #- 1/(-1)=1# (and as already noted passes through (6,7)#

A slope of #1# implies a #Deltax:Deltay# ratio of #1:1# (with a corresponding hypotenuse of #sqrt(2)#).

A point on the perpendicular bisector a distance of #8sqrt(2)# from #(6,7)# will be
either at #(6-8,7-8)=(-2,-1)#
or at #(6+8,7+8) = (14,15)#
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