Find the Area of the Triangle (0,5) , (3,4) , (2,0)

$(0, 5)$ , $(3, 4)$ , $(2, 0)$

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.

$\frac{| 0 (B_{y} - C_{y}) + B_{x} (C_{y} - 5) + C_{x} (5 - B_{y}) |}{2}$

Substitute the values of the second point into the formula.

$\frac{| 0 (4 - C_{y}) + 3 (C_{y} - 5) + C_{x} (5 - 1 \cdot 4) |}{2}$

Substitute the values of the final point into the formula.

$\frac{| 0 (4 + 0) + 3 (0 - 5) + 2 (5 - 1 \cdot 4) |}{2}$

Step 3

Simplify the numerator.

Add $4$ and $0$ .

$\frac{| 0 \cdot 4 + 3 (0 - 5) + 2 (5 - 1 \cdot 4) |}{2}$

Multiply $0$ by $4$ .

$\frac{| 0 + 3 (0 - 5) + 2 (5 - 1 \cdot 4) |}{2}$

Subtract $5$ from $0$ .

$\frac{| 0 + 3 \cdot - 5 + 2 (5 - 1 \cdot 4) |}{2}$

Multiply $3$ by $- 5$ .

$\frac{| 0 - 15 + 2 (5 - 1 \cdot 4) |}{2}$

Multiply $- 1$ by $4$ .

$\frac{| 0 - 15 + 2 (5 - 4) |}{2}$

Subtract $4$ from $5$ .

$\frac{| 0 - 15 + 2 \cdot 1 |}{2}$

Multiply $2$ by $1$ .

$\frac{| 0 - 15 + 2 |}{2}$

Subtract $15$ from $0$ .

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

Add $- 15$ and $2$ .

$\frac{| - 13 |}{2}$

The absolute value is the distance between a number and zero. The distance between $- 13$ and $0$ is $13$ .

$\frac{13}{2}$