Question

Find the area of a triangle whose vertices are located at (-2,5), (-4, -3), and (3,1). Evaluate the determinant using diagonals.

Answer 1

Answer:

The area of the triangle is 48 unit²

Step-by-step explanation:

The given vertices of the triangle are;

(-2, 5), (-4, -3), and (3, 1)

The formula for finding the area of a triangle with given coordinates of the vertices is as follows;

$\Delta = \dfrac{1}{2}\times \begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} = \dfrac{1}{2}\times \left | x_1\cdot y_2 - x_2\cdot y_1 + x_2\cdot y_3 - x_3\cdot y_2 + x_3\cdot y_1 - x_1\cdot y_3\right |$

Substituting gives;

$\Delta = \dfrac{1}{2}\times \begin{vmatrix}-2& 5 & 1\\ -4 & -3 & 1\\ 3 & 1 & 1\end{vmatrix} \\\Delta = \dfrac{1}{2}\times \left | (-2)\times (-3) - ((-4)\times 5) + (-4)\times 1 - 3\times (-3) + 3 \times 5 - (-2)\times 1\right | \\\Delta = \dfrac{1}{2}\times \left | 6 +20 -4 +9 + 15+2\right | = 48$

The area of the triangle = 48 unit².