Question

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?

Answer 1

(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.