An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(4 ,3 )# to #(8 ,9 )# and the triangle's area is #36 #, what are the possible coordinates of the triangle's third corner?

Question

1169 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-08T09:27:06

The coordinates of the third corner are #(14.3,0.5)# or #(-2.3,11.5)#

Let the coordinates of the third corner be #=(x,y)#

The mid point of side #A# is #=((8+4)/2,(9+3)/2)=(6,6)#

We apply Pythagoras' theorem to the triangles

#(x-4)^2+(y-3)^2=(6-4)^2+(6-3)^2+(x-6)^2+(y-6)^2#

#x^2-8x+16+y^2-6y+9=4+9+x^2-12x+36+y^2-12y+36#

#-8x-6y+25=85-12x-12y#

#4x+6y=60#

#2x+3y=30#...........................#(1)#

We now write the second equation

The area is

#a=1/2*b*h#

#1/2(sqrt((8-4)^2+(9-3)^2))(sqrt((x-6)^2+(y-6)^2))=36#

#sqrt52*(sqrt((x-6)^2+(y-6)^2))=72#

#52((x-6)^2+(y-6)^2)=72^2=5184#

#(x-6)^2+(y-6)^2=99.7#.............................#(2)#

Solving for #x# and #y# in equations #(1)# and #(2)#

#(x-6)^2+((30-2x)/3-6)^2=99.7#

#x^2-12x+36+(144-48x+4x^2)/9=99.7#

#9x^2-108x+324+144-48x+4x^2=897.3#

#13x^2-156x-429.3=0#

We solve this quadratic equation

#x=(156+-sqrt(156^2-4*13*(-429.3)))/(26)#

#x_1=(156+216)/26=14.3#

#x_2=(156-216)/26=-2.3#

#y_1=(30-2*14.3)/3=0.5#

#y_2=(30+2*2.3)/3=11.5#