Find the Area of the Triangle (-3,4) , (4,4) , (3,-3)

$(- 3, 4)$ , $(4, 4)$ , $(3, - 3)$

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.

$\frac{| - 3 (B_{y} - C_{y}) + B_{x} (C_{y} - 4) + C_{x} (4 - B_{y}) |}{2}$

Substitute the values of the second point into the formula.

$\frac{| - 3 (4 - C_{y}) + 4 (C_{y} - 4) + C_{x} (4 - 1 \cdot 4) |}{2}$

Substitute the values of the final point into the formula.

$\frac{| - 3 (4 + 3) + 4 (- 3 - 4) + 3 (4 - 1 \cdot 4) |}{2}$

Step 3

Simplify the numerator.

Add $4$ and $3$ .

$\frac{| - 3 \cdot 7 + 4 (- 3 - 4) + 3 (4 - 1 \cdot 4) |}{2}$

Multiply $- 3$ by $7$ .

$\frac{| - 21 + 4 (- 3 - 4) + 3 (4 - 1 \cdot 4) |}{2}$

Subtract $4$ from $- 3$ .

$\frac{| - 21 + 4 \cdot - 7 + 3 (4 - 1 \cdot 4) |}{2}$

Multiply $4$ by $- 7$ .

$\frac{| - 21 - 28 + 3 (4 - 1 \cdot 4) |}{2}$

Multiply $- 1$ by $4$ .

$\frac{| - 21 - 28 + 3 (4 - 4) |}{2}$

Subtract $4$ from $4$ .

$\frac{| - 21 - 28 + 3 \cdot 0 |}{2}$

Multiply $3$ by $0$ .

$\frac{| - 21 - 28 + 0 |}{2}$

Add $- 21 - 28$ and $0$ .

$\frac{| - 21 - 28 |}{2}$

Subtract $28$ from $- 21$ .

$\frac{| - 49 |}{2}$

The absolute value is the distance between a number and zero. The distance between $- 49$ and $0$ is $49$ .

$\frac{49}{2}$