An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(1 ,4 )# to #(5 ,8 )# and the triangle's area is #27 #, what are the possible coordinates of the triangle's third corner?

1 Answer
May 29, 2018

(9.75, -0.75) or (-3.75, 12.75)


First picture side A, common to two possible isosceles triangles, from (1,4) to (5,8). By Pythagoras we see that this side has length #4sqrt(2)#.

Now draw the line perpendicular to side A which also bisects side A, up until it reaches where sides B and C of the isosceles triangle meet; this new line is the height of the triangle.

Now you will realise that the area of the triangle, 27, is equal to #1/2bh = 1/2*4sqrt(2)*h#. Solving for #h#, we see that #h = (27sqrt(2))/4#.

The final thing to note is that since this line is perpendicular to side A, which has a gradient of 1, this line has a gradient of -1. Hence, the horizontal and vertical components of the triangle connecting the midpoint of A and the other point of the isosceles triangle are equal in magnitude. Hence we can set up the equation #sqrt(2x^2)=(27sqrt(2))/4#, which when solved gives #x=+-27/4.#

Now all that is left is to add this value of #x# to the midpoint #(3,6)# of side A, taking care with the signs, to obtain the possible coordinates of the final point.

The point of the possible triangle to the bottom right of the midpoint is #(3+27/4,6-27/4) = (9.75, -0.75)# and the point to the top left of the midpoint is #(3-27/4, 6+27/4) = (-3.75, 12.75)#, both as required.