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?

A(-3,5) B(3,7) C(2,3)

Answer 1

Are you sure that's right

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 defined by three points in a Cartesian coordinate system. The formula is given by:

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

We denote the points as follows:
- \( A(x_1, y_1) = (-3, 5) \)
- \( B(x_2, y_2) = (3, 7) \)
- \( C(x_3, y_3) = (2, 3) \)

Now, we substitute the coordinates into the formula:

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

Calculating each term separately:
1. First term: \( -3(7 - 3) = -3 \times 4 = -12 \)
2. Second term: \( 3(3 - 5) = 3 \times (-2) = -6 \)
3. Third term: \( 2(5 - 7) = 2 \times (-2) = -4 \)

Now, we can sum these values:

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

Thus, the area of triangle ABC is \( \boxed{11} \) square units.

Answer 3

Let’s carefully go through the calculations together to ensure correctness.

We have the vertices of the triangle \( A(-3, 5) \), \( B(3, 7) \), and \( C(2, 3) \). To find the area, we will use the formula for the area of a triangle given by three vertices:

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

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

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

Calculating each component:

1. **For \( A \)**:
\[
-3(7 - 3) = -3 \times 4 = -12
\]

2. **For \( B \)**:
\[
3(3 - 5) = 3 \times (-2) = -6
\]

3. **For \( C \)**:
\[
2(5 - 7) = 2 \times (-2) = -4
\]

Now, we sum these results:
\[
-12 - 6 - 4 = -22
\]

Thus, we take the absolute value:
\[
\left| -22 \right| = 22
\]

Then we apply the final step:
\[
\text{Area} = \frac{1}{2} \times 22 = 11
\]

The area of triangle ABC is indeed \( \boxed{11} \) square units. The calculations are confirmed to be correct. Thank you for your patience.