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)

Explanation:

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.