Pre-Algebra Examples

$(1, 1)$ , $(2, 2)$ , $(3, 0)$

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.

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

Step 2.2

Substitute the values of the second point into the formula.

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

Step 2.3

Substitute the values of the final point into the formula.

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

Step 3

Simplify the numerator.

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Step 3.1

Multiply $2 + 0$ by $1$ .

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

Step 3.2

Add $2$ and $0$ .

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

Step 3.3

Subtract $1$ from $0$ .

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

Step 3.4

Multiply $2$ by $- 1$ .

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

Step 3.5

Multiply $- 1$ by $2$ .

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

Step 3.6

Subtract $2$ from $1$ .

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

Step 3.7

Multiply $3$ by $- 1$ .

$\frac{| 2 - 2 - 3 |}{2}$

Step 3.8

Subtract $2$ from $2$ .

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

Step 3.9

Subtract $3$ from $0$ .

$\frac{| - 3 |}{2}$

Step 3.10

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$\frac{3}{2}$

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