Find the Area of the Triangle (3,1) , (3,-2) , (-2,-2)

$(3, 1)$ , $(3, - 2)$ , $(- 2, - 2)$

Step 1

Use the formula to find the area of the triangle with $3$ points $(A_{x}, A_{y})$ , $(B_{x}, B_{y})$ , $(C_{x}, C_{y})$ .

$Area = \frac{| A_{x} (B_{y} - C_{y}) + B_{x} (C_{y} - A_{y}) + C_{x} (A_{y} - B_{y}) |}{2}$

Step 2

Substitute the points into the formula.

Substitute the values of the first point into the formula.

$\frac{| 3 (B_{y} - C_{y}) + B_{x} (C_{y} - 1) + C_{x} (1 - B_{y}) |}{2}$

Substitute the values of the second point into the formula.

$\frac{| 3 (- 2 - C_{y}) + 3 (C_{y} - 1) + C_{x} (1 - - 2) |}{2}$

Substitute the values of the final point into the formula.

$\frac{| 3 (- 2 + 2) + 3 (- 2 - 1) - 2 (1 - - 2) |}{2}$

Step 3

Simplify the numerator.

Add $- 2$ and $2$ .

$\frac{| 3 \cdot 0 + 3 (- 2 - 1) - 2 (1 - - 2) |}{2}$

Multiply $3$ by $0$ .

$\frac{| 0 + 3 (- 2 - 1) - 2 (1 - - 2) |}{2}$

Subtract $1$ from $- 2$ .

$\frac{| 0 + 3 \cdot - 3 - 2 (1 - - 2) |}{2}$

Multiply $3$ by $- 3$ .

$\frac{| 0 - 9 - 2 (1 - - 2) |}{2}$

Multiply $- 1$ by $- 2$ .

$\frac{| 0 - 9 - 2 (1 + 2) |}{2}$

Add $1$ and $2$ .

$\frac{| 0 - 9 - 2 \cdot 3 |}{2}$

Multiply $- 2$ by $3$ .

$\frac{| 0 - 9 - 6 |}{2}$

Subtract $9$ from $0$ .

$\frac{| - 9 - 6 |}{2}$

Subtract $6$ from $- 9$ .

$\frac{| - 15 |}{2}$

The absolute value is the distance between a number and zero. The distance between $- 15$ and $0$ is $15$ .

$\frac{15}{2}$