Question

Amaury draws an acute triangle on the coordinate grid. He finds the area of the triangle by drawing a rectangle that circumscribes the triangle and forming right triangles. How can Amaury find the area of the triangle? What is the area of the triangle in square units?
Point A: (-3, 5)
Point B: (3, 7)
Point C: (2, 3)

Answer 1

Are you sure that's the right answer?

Answer 2

To find the area of triangle ABC with vertices at points A(-3, 5), B(3, 7), and C(2, 3), we can use the formula for the area of a triangle given by the coordinates of its vertices:

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

Here, let:
- \( A(x_1, y_1) = (-3, 5) \)
- \( B(x_2, y_2) = (3, 7) \)
- \( C(x_3, y_3) = (2, 3) \)

Substituting the coordinates into the formula, we have:

\[
\text{Area} = \frac{1}{2} \left| (-3)(7 - 3) + 3(3 - 5) + 2(5 - 7) \right|
\]

Calculating each term step-by-step:

1. Calculate \( -3(7 - 3) = -3 \times 4 = -12 \)
2. Calculate \( 3(3 - 5) = 3 \times -2 = -6 \)
3. Calculate \( 2(5 - 7) = 2 \times -2 = -4 \)

Now, substituting back into the area formula:

\[
\text{Area} = \frac{1}{2} \left| -12 - 6 - 4 \right| = \frac{1}{2} \left| -22 \right| = \frac{1}{2} \times 22 = 11
\]

Thus, the area of triangle ABC is:

\[
\boxed{11} \text{ square units}
\]