Find the area of a triangle A (11: 5) B (3: 8) C (6: -7)

By the coordinates of the vertices of the triangle, we determine the lengths of its sides.

The length of the segment AB = √ (X2 – X1) ^ 2 + (Y2 – Y1) ^ 2 = √ (3 – 11) ^ 2 + (8 – 5) ^ 2 = √ (64 + 9) = √73 = 8, 54.

The length of the segment AC = √ (6 – 11) ^ 2 + (-7 – 5) ^ 2 = √ (25 + 144) = √169 = 13.

The length of the segment BC = √ (6 – 3) ^ 2 + (-7 – 8) ^ 2 = √ (9 + 225) = √ 234 = 15.3.

Let’s define the semiperimeter of the triangle.

P = (8.54 + 13 + 15.3) / 2 = 18.42.

By Heron’s theorem:

S = √18.42 * (18.42 – 8.54) * (18.42 – 13) * (18.42 – 15.3) = √18.42 * 9.88 * 5.42 * 3, 12 = 55.5 units ^ 2.

Answer: the area of the triangle is 55.5 units ^ 2.