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 ,5 )` and the triangle's area is `32`, what are the possible coordinates of the triangle's third corner?

Answer 1

Coordinates of vertex A `color(brown)(12.74, -0.68)`

Explanation:

Given : B (4,9), C (1,5), `A_t = 32`, AB = AC, `AN (h) _|_` bisector of BC

To find A coordinates.

Coordinates of N (4+1)/2, (9+5)/2 = N (5/2, 7)

`BC = a = sqrt((1-4)^2 + (5-9)^2) = 5`

`A_t = 32 = (1/2) a * h = (1/2) * 5 * h`

`h = 12.8`

Slope of BC `m_a = (5-9) / (1 - 4) = 4/3`

Slope of `AN _|_ BC = m_h = -1 / m_a = -3/4`

Equation of AN

`y - 7 = -(3/4) (x - (5/2))`

`y - 7 = -(3/8)(2x - 5)`

`8y - 56 = -6x + 15`

`8y + 6x = 71` Eqn (1)

AN = h = 12.8 = sqrt((x-(5/2))^2 + (y-7)^2)#

`(y-7)^2 + (x-(5/2))^2 = 12.8^2` Eqn (2)

Solving equations (1), (2), we get the coordinates of vertex A

`A (12.74, -0.68)`

Verification :

`AB = c = sqrt((4-12.74)^2 + (9+0.68)^2) = 13.0419`

`AC = b = sqrt((1-12.74)^2 + (5+0.68)^2) = 13.0419`

`b = c = 13.0419` since it is an isosceles triangle