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 ,5 )` to `(8 ,1 )` and the triangle's area is `15`, what are the possible coordinates of the triangle's third corner?

Answer 1

Coordinates of the third vertex is `(187/9), (80/3)`

Explanation:

Area = 15

`a = sqrt ( (8-2)^2 + (1-5)^2) = sqrt 52 = 2sqrt 13`

`Area = (1/2) a * h`

`15 = (1/2) * 2 sqrt 13 * h`

`h = 15/ ( sqrt 13)`

`tan theta = h / (a/2) = (15/sqrt13) / ((2 sqrt 13)/2) = 15/13`

Slope of line AB is `tan theta = 15/13`
Equation of line AB is given by
`y - 5 = (15/13) * ( x - 2)`
`13y - 65 = 15x - 30`
`15x - 13y = —35`. Eqn (1)

Slope of base `m = (1-5)/(8-2) = -2/3`
Slope of altitude `m_1 = -(1/m) = 3/2`

Midpoint of base a is `(8+2)/2, (1+5)/2= (5, 3)`

Equation of altitude and passing through point A and having slope 3/2 is
`y - 3 = (3/2) * (x - 5)`
`2y - 6 = 3x - 15`
`3x - 2y = 9` Eqn (2)

Solving Eqns (1) & (2), we get coordinates of point A.

`x = (187/9), y =( 80/3)`

Coordinates of the third corner A is `(187/9, 80/3)`