Question

An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from `(7 ,5 )` to `(8 ,2 )` and the triangle's area is `27`, what are the possible coordinates of the triangle's third corner?

Answer 1

The coordinates are `(23.7,8.9)` and `(-8.7,-1.9)`

Explanation:

The length of side `A=sqrt((7-8)^2+(5-2)^2)=sqrt10`

Let the height of the triangle be `=h`

The area of the triangle is

`1/2*sqrt10*h=27`

The altitude of the triangle is `h=(27*2)/sqrt10=54/sqrt10`

The mid-point of `A` is `(15/2,7/2)`

The gradient of `A` is `=(2-5)/(8-7)=-3`

The gradient of the altitude is `=1/3`

The equation of the altitude is

`y-7/2=1/3(x-15/2)`

`y=1/3x-5/2+7/2=1/3x+1`

The circle with equation

`(x-15/2)^2+(y-7/2)^2=54^2/10=291.6`

The intersection of this circle with the altitude will give the third corner.

`(x-15/2)^2+(1/3x+1-7/2)^2=291.6`

`x^2-15x+225/4+1/9x^2-5/3x+25/4=291.6`

`1.11x^2-16.7x-229.1=0`

We solve this quadratic equation

`x=(16.7+-sqrt(16.7^2+4*1.11*229.1))/(2*1.11)`

`=(16.7+-36)/2.22`

`x_1=23.74`

`x_2=-8.69`

The points are `(23.7,9.65)` and `(-8.7,-1.9)`

graph{(y-1/3x-1)((x-7.5)^2+(y-3.5)^2-291.6)((x-7)^2+(y-5)^2-0.05)((x-8)^2+(y-2)^2-0.05)(y-5+3(x-7))=0 [-12, 28, -10, 10]}