Question

An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from `(4 ,9 )` to `(8 ,5 )` and the triangle's area is `64`, what are the possible coordinates of the triangle's third corner?

Answer 1

The third corner will be at either `(-2,-1)` or `(14,15)`

Explanation:

Length of side A = `sqrt((8-4)^2+(5-9)^2) = 4sqrt(2)`

Height of triangle ABC relative to A `= ("area")/("length side A")`
`color(white)("XXX")64/(4sqrt(2))= 8sqrt(2)`

Note that since `abs(B)=abs(C)` the height relative to `A` is the length of the perpendicular bisector of `A`

Mid-point of `A` is `((4+8)/2,(9+5)/2)=(6,7)`

`A` has a slope of `(Deltay)/(Deltax)=(9-5)/(4-8)=-1`

`rArr` the perpendicular bisector has a slope of `- 1/(-1)=1` (and as already noted passes through (6,7)#

A slope of `1` implies a `Deltax:Deltay` ratio of `1:1` (with a corresponding hypotenuse of `sqrt(2)`).

A point on the perpendicular bisector a distance of `8sqrt(2)` from `(6,7)` will be
either at `(6-8,7-8)=(-2,-1)`
or at `(6+8,7+8) = (14,15)`