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

Let's assume the vertices of a triangle to be:

A = (x₁, y₁) = (-5, 7)

B = (x₂, y₂) = (-4, -5)

C = (x₃, y₃) = (4, 5)

Area of triangle ABC = $\frac{1}{2}\left|x_1(y_2 – y_3) + x_2(y_{3 }– y_1) + x_3(y_{1 }– y_2)\right|$

Area of triangle ABC = $\frac{1}{2}\left|-5(-5-5)+(-4)(5-7)+\left(4\right)\left(7-\left(-5\right)\right)\right|$

Area of triangle ABC = $\frac{1}{2}\left|50+8+48\right|$

Area of triangle ABC = 53 sq. units