Question

An isosceles triangle has sides `a, b` and `c` with sides `b` and `c` being equal in length. If side `a` goes from `(2 ,9 )` to `(8 ,5 )` and the triangle's area is `64`, what are the possible coordinates of the triangle's third corner?

Answer 1

`(-63/13, -101/13) or (193/13, 283/13)`

Explanation:

Side `a` is `(2,9)` to `(8,5)`

Let's do this directly. We have

`a^2 =(8-2)^2+(5-9)^2=36+16=52`

Area `S = 1/2 a h` or

`h = {2S}/a = {2(64)}/\sqrt{52} = 128/sqrt{52}`

The other vertex is along the perpendicular bisector of side `a`. The direction vector for side as is `(8-2,5-9)=(6,-4)` so the perpendicular through the midpoint is

`(x,y) = ( {2+8}/2, {9+5}/2) + t(4,6) = (5,7)+t(4,6)`

We need `|t(4,6)|=h`

`t = pm h/|(4.6)| = pm (128/sqrt{52})/sqrt{52} = pm 128/52 = pm 32/13`

Two choices for the third vertex:

#(x,y) = (5,7) pm (32/13) (4,6) = (-63/13, -101/13) or (193/13, 283/13)#

Let's check one:

`b^2 = (-63/13 - 2)^2 + (-101/13 - 9)^2 = 4265/13`

`c^2 = (-63/13 - 8)^2 + (-101/13 - 5)^2 = 4265/13 quad sqrt`

Area `S`

`16S^2 = 4(52)(4265/13) - (52)^2`

`S = 64 quad sqrt`