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 #27 #, what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2017-07-01T14:11:19

The coordinates are #(23.7,8.9)# and #(-8.7,-1.9)#

The length of side #A=sqrt((7-8)^2+(5-2)^2)=sqrt10#

Let the height of the triangle be #=h#

The area of the triangle is

#1/2*sqrt10*h=27#

The altitude of the triangle is #h=(27*2)/sqrt10=54/sqrt10#

The mid-point of #A# is #(15/2,7/2)#

The gradient of #A# is #=(2-5)/(8-7)=-3#

The gradient of the altitude is #=1/3#

The equation of the altitude is

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

#y=1/3x-5/2+7/2=1/3x+1#

The circle with equation

#(x-15/2)^2+(y-7/2)^2=54^2/10=291.6#

The intersection of this circle with the altitude will give the third corner.

#(x-15/2)^2+(1/3x+1-7/2)^2=291.6#

#x^2-15x+225/4+1/9x^2-5/3x+25/4=291.6#

#1.11x^2-16.7x-229.1=0#

We solve this quadratic equation

#x=(16.7+-sqrt(16.7^2+4*1.11*229.1))/(2*1.11)#

#=(16.7+-36)/2.22#

#x_1=23.74#

#x_2=-8.69#

The points are #(23.7,9.65)# and #(-8.7,-1.9)#

graph{(y-1/3x-1)((x-7.5)^2+(y-3.5)^2-291.6)((x-7)^2+(y-5)^2-0.05)((x-8)^2+(y-2)^2-0.05)(y-5+3(x-7))=0 [-12, 28, -10, 10]}