# 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)

Nov 21, 2016

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

#### Explanation:

$\overline{A B} = \left(- 1 - - 2\right) \hat{i} + \left(1 - 4\right) \hat{j} = \hat{i} - 3 \hat{j}$

$\overline{A C} = \left({C}_{x} - - 2\right) \hat{i} + \left({C}_{y} - 4\right) \hat{j}$

$\overline{B C} = \left({C}_{x} - - 1\right) \hat{i} + \left({C}_{y} - 1\right) \hat{j}$

$\overline{A B} \cdot \overline{A C} = \left({C}_{x} + 2\right) - 3 \left({C}_{y} - 4\right)$

$\overline{A B} \cdot \overline{B C} = \left({C}_{x} + 1\right) - 3 \left({C}_{y} - 1\right)$

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

$C = \left(2 , 1\right)$

$\overline{A B} \cdot \overline{A C} = \left(2 + 2\right) - 3 \left(1 - 4\right) = 4 + 9 = 13$

$\overline{A B} \cdot \overline{B C} = \left(2 + 1\right) - 3 \left(1 - 1\right) = 3 + 0 = 3$

Not a right triangle.

$C = \left(0 , 4\right)$

$\overline{A B} \cdot \overline{A C} = \left(0 + 2\right) - 3 \left(4 - 4\right) = 2 + 0 = 2$

$\overline{A B} \cdot \overline{B C} = \left(0 + 1\right) - 3 \left(4 - 1\right) = 1 - 9 = - 8$

Not a right triangle.

$C = \left(2 , 2\right)$

$\overline{A B} \cdot \overline{A C} = \left(2 + 2\right) - 3 \left(2 - 4\right) = 4 + 6 = 10$

$\overline{A B} \cdot \overline{B C} = \left(2 + 1\right) - 3 \left(2 - 1\right) = 3 - 3 = 0$

This is a right triangle.