Algebra Examples

$(- 4, - 5)$ , $(- 3, 2)$ and $(9, 0)$

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.

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Step 2.1

Substitute the values of the first point into the formula.

$\frac{| - 4 (B_{y} - C_{y}) + B_{x} (C_{y} + 5) + C_{x} (- 5 - B_{y}) |}{2}$

Step 2.2

Substitute the values of the second point into the formula.

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

Step 2.3

Substitute the values of the final point into the formula.

$\frac{| - 4 (2 + 0) - 3 (0 + 5) + 9 (- 5 - 1 \cdot 2) |}{2}$

Step 3

Simplify to find the area of the triangle.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Add $2$ and $0$ .

$\frac{| - 4 \cdot 2 - 3 (0 + 5) + 9 (- 5 - 1 \cdot 2) |}{2}$

Step 3.1.2

Multiply $- 4$ by $2$ .

$\frac{| - 8 - 3 (0 + 5) + 9 (- 5 - 1 \cdot 2) |}{2}$

Step 3.1.3

Add $0$ and $5$ .

$\frac{| - 8 - 3 \cdot 5 + 9 (- 5 - 1 \cdot 2) |}{2}$

Step 3.1.4

Multiply $- 3$ by $5$ .

$\frac{| - 8 - 15 + 9 (- 5 - 1 \cdot 2) |}{2}$

Step 3.1.5

Multiply $- 1$ by $2$ .

$\frac{| - 8 - 15 + 9 (- 5 - 2) |}{2}$

Step 3.1.6

Subtract $2$ from $- 5$ .

$\frac{| - 8 - 15 + 9 \cdot - 7 |}{2}$

Step 3.1.7

Multiply $9$ by $- 7$ .

$\frac{| - 8 - 15 - 63 |}{2}$

Step 3.1.8

Subtract $15$ from $- 8$ .

$\frac{| - 23 - 63 |}{2}$

Step 3.1.9

Subtract $63$ from $- 23$ .

$\frac{| - 86 |}{2}$

Step 3.1.10

The absolute value is the distance between a number and zero. The distance between $- 86$ and $0$ is $86$ .

$\frac{86}{2}$

Step 3.2

Divide $86$ by $2$ .

$43$