Find the area of a triangle ABC if: A (0; -2), B (5; -2), C (5; 0).

Let us find the lengths of the sides of this triangle using the formula for the distance between two points A and B on the coordinate plane with coordinates A (x1; y1) and B (x2; y2):

| AB | = √ ((x1 – x2) ^ 2 + (y1 – y2) ^ 2).

| AB | = √ ((0 – 5) ^ 2 + (-2 – (-2)) ^ 2) = √ (5 ^ 2 + 0 ^ 2) = √5 ^ 2 = 5;

| AC | = √ ((0 – 5) ^ 2 + (-2 – 0) ^ 2) = = √ (5 ^ 2 + 2 ^ 2) = √ (25 + 4) = √29;

| BC | = √ ((5 – 5) ^ 2 + (-2 – 0) ^ 2) = √ (0 ^ 2 + 2 ^ 2) = √2 ^ 2 = 2.

Since 5 ^ 2 + 2 ^ 2 = (√29) ^ 2, then for this triangle the relation | AB | ^ 2 + | BC | ^ 2 = | AC | ^ 2 is fulfilled.

Consequently, this triangle is right-angled with legs AB and BC and its area S is equal to half the product of the lengths of its legs:

S = | AB | * | BC | / 2 = 5 * 2/2 = 5.

Answer: the area of ​​this triangle is 5.