Question

In ∆ABC the coordinates vertices A and B are A (-2, 4) and B (-1, 1). For each of the given coordinates of vertex C, is ∆ ABC a right triangle?

C(-2, 1)

C(0,4)

C(2,2)

Answer 1

The only right triangle is `C = (2,2)`. Found by the dot-product.

Explanation:

`bar(AB) = (-1 - -2)hati + (1 - 4)hatj = hati - 3hatj`

`bar(AC) = (C_x - -2)hati + (C_y - 4)hatj`

`bar(BC) = (C_x - -1)hati + (C_y - 1)hatj`

`bar(AB)*bar(AC) = (C_x + 2) - 3(C_y - 4)`

`bar(AB)*bar(BC) = (C_x + 1) - 3(C_y - 1)`

The triangle will be a right triangle if either dot-product is zero:

`C = (2, 1)`

`bar(AB)*bar(AC) = (2 + 2) - 3(1 - 4) = 4 + 9 = 13`

`bar(AB)*bar(BC) = (2 + 1) - 3(1 - 1) = 3 + 0 = 3`

Not a right triangle.

`C = (0, 4)`

`bar(AB)*bar(AC) = (0 + 2) - 3(4 - 4) = 2 + 0 = 2`

`bar(AB)*bar(BC) = (0 + 1) - 3(4 - 1) = 1 - 9 = -8`

Not a right triangle.

`C = (2, 2)`

`bar(AB)*bar(AC) = (2 + 2) - 3(2 - 4) = 4 + 6 = 10`

`bar(AB)*bar(BC) = (2 + 1) - 3(2 - 1) = 3 - 3 = 0`

This is a right triangle.