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

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 its vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):

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

Substituting in the coordinates of points A, B, and C:

• \(A(-3, 5)\) corresponds to \((x_1, y_1) = (-3, 5)\)
• \(B(3, 7)\) corresponds to \((x_2, y_2) = (3, 7)\)
• \(C(2, 3)\) corresponds to \((x_3, y_3) = (2, 3)\)

Now we substitute these values into the formula:

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

Calculating each term step-by-step:

1. The first term: \(-3(7 - 3) = -3(4) = -12\)
2. The second term: \(3(3 - 5) = 3(-2) = -6\)
3. The third term: \(2(5 - 7) = 2(-2) = -4\)

Putting it all together:

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

Thus, the area of triangle ABC is

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