Algebra Examples

(- 4, - 5)

$(- 4, - 5)$ ,

(- 3, 2)

$(- 3, 2)$ and

(9, 0)

$(9, 0)$

Step 1

Use the formula to find the area of the triangle with

3

$3$ points

(A_{x}, A_{y})

$(A_{x}, A_{y})$ ,

(B_{x}, B_{y})

$(B_{x}, B_{y})$ ,

(C_{x}, C_{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}

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

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}

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

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}

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

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}

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

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

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

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

$2$ and

0

$0$ .

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

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

Step 3.1.2

Multiply

- 4

$- 4$ by

2

$2$ .

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

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

Step 3.1.3

Add

0

$0$ and

5

$5$ .

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

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

Step 3.1.4

Multiply

- 3

$- 3$ by

5

$5$ .

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

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

Step 3.1.5

Multiply

- 1

$- 1$ by

2

$2$ .

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

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

Step 3.1.6

Subtract

2

$2$ from

- 5

$- 5$ .

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

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

Step 3.1.7

Multiply

9

$9$ by

- 7

$- 7$ .

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

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

Step 3.1.8

Subtract

15

$15$ from

- 8

$- 8$ .

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

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

Step 3.1.9

Subtract

63

$63$ from

- 23

$- 23$ .

\frac{| - 86 |}{2}

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

Step 3.1.10

The absolute value is the distance between a number and zero. The distance between

- 86

$- 86$ and

0

$0$ is

86

$86$ .

\frac{86}{2}

$\frac{86}{2}$

\frac{86}{2}

$\frac{86}{2}$

Step 3.2

Divide

86

$86$ by

2

$2$ .

43

$43$

43

$43$