if the coordinates of a triangles points are; (0, 3), (10, 6), and (6, 3), what is the triangle's area in square units?

Answer 1

To find the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), we can use the formula:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of the triangle's points:

• \((x_1, y_1) = (0, 3)\)
• \((x_2, y_2) = (10, 6)\)
• \((x_3, y_3) = (6, 3)\)

Let's plug these values into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 0(6 - 3) + 10(3 - 3) + 6(3 - 6) \right| \]

Calculating each term step-by-step:

1. \(0(6 - 3) = 0\)
2. \(10(3 - 3) = 10(0) = 0\)
3. \(6(3 - 6) = 6(-3) = -18\)

Now, combining the terms in the absolute value:

\[ \text{Area} = \frac{1}{2} \left| 0 + 0 - 18 \right| = \frac{1}{2} \left| -18 \right| = \frac{1}{2} \times 18 = 9 \]

Therefore, the area of the triangle is \(\boxed{9}\) square units.