Question

Find the area of the triangle whose vertices are \( (0,4) \), \( (3,6) \), and \( (-8,-2) \).

Hint:

Always take the absolute value when calculating area using determinants.

Answer 1

The area of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \] Substituting values:
\[ A = \frac{1}{2} \left| 0(6 + 2) + 3(-2 - 4) + (-8)(4 - 6) \right|. \] \[ = \frac{1}{2} \left| 0 + 3(-6) + (-8)(-2) \right|. \] \[ = \frac{1}{2} \left| -18 + 16 \right| = \frac{1}{2} \times 2 = 1. \]