An isosceles triangle has sides #a,# #b,# and #c# with sides #a# and #c# being equal in length. If side #b# goes from #(5 ,1 )# to #(3 ,2 )# and the triangle's area is #8 #, what are the possible coordinates of the triangle's third vertex?

1 Answer
Dean R.
Jun 9, 2018

#(23/2, 19/2) or (-9/2, -13/2)#

Explanation:

We have #A(4,1)# and #C(3,2)# and unknown third vertex #B(x,y),#

For AC we have direction vector #C-A=(3-4, 2-1)=(-1,1).# The perpendicular has direction vector given by swapping the coordinates and negating one of them, #(1,1)#.

The foot of the bisector is the midpoint of #AC#, #F({4+3}/2, {1+2}/2) = F(7/2, 3/2)#

#B# lies along the perpendicular bisector of AC, so through F.

# (x,y) = F + t(1,1) = (t+ 7/2, t +3/2)#

The area is #S=8.# We can use the Shoelace Theorem:

#S=8= 1/2 | 4(t+3/2) - (t+7/2) + 3(1)-2(4) + (t+7/2)(2) - (t+3/2)(3) |#

#pm 16 = 2t#

#t = pm 8#

# (x,y) = (t+ 7/2, t +3/2)#

#(x,y) = (23/2, 19/2) or (-9/2, -13/2)#

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