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 `(1 ,0 )` and the triangle's area is `24`, what are the possible coordinates of the triangle's third corner?

Answer 1

Coordinates of the third corner A (4.5713, 3.8096)

Explanation:

`BC = sqrt((4-1)^2 + (9-0)^2) = 9.4868`

Area of triangle `A_t = 24 = (1/2)* BC * AD = (1/2) * 9.4868 * h`

`h = (2*24) / 9.4868 = 5.06`

Slope of BC `m_(BC) = (0-9) / (1-4) = 3`

Slope of AD `m_(AD) = -(1/m_(BC)) = -1/3`

Coordinates of point D is
`= (4+1)/2, (9+0)/2 = (2.5, 4.5)`

Equation of AD is
`(y - 4.5) = (-1/3)(x - 2.5)`

`3y + x = 16` Eqn (1)

`tan C = m_(CA) = (AD) / (CD) = 5.06 / (9.4868/2) = 1.0667`

Equation of CA is
`y - 0 = 1.0667(x - 1)`

`y - 1.0667 x = - 1.0667` Eqn (2)

Solving Equations (1) & (2), we get the coordinates of A

`x = (4.5713), y = (3.8096)`