The area of the triangle ABC with the vertices A(–5, 7), B(–4, –5) and C(4, 5) is

To find the area of a triangle with vertices at three points (x1, y1), (x2, y2), and (x3, y3), you can use the formula:

Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Here, the vertices of the triangle ABC are A(-5, 7), B(-4, -5), and C(4, 5). So, x1 = -5, y1 = 7, x2 = -4, y2 = -5, x3 = 4, and y3 = 5.

Substitute these values into the formula:

Area = 1/2 * |-5(-5 - 5) - 4(5 - 7) + 4(7 - (-5))| = 1/2 * |-5(-10) - 4(-2) + 4(12)| = 1/2 * |50 + 8 + 48| = 1/2 * 106 = 53 square units

So, the area of the triangle ABC is 53 square units.