Prove that the triangle with vertices A (2; 3) B (-1; 1) C (3; -4) is isosceles.

Let’s use 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)

and find the lengths of the sides of this triangle.

| AB | = √ ((2 – (-1)) ^ 2 + (3 – 1) ^ 2) = √ (3 ^ 2 + 2 ^ 2) = √ (9 + 4) = √13;

| BC | = √ ((3 – (-1)) ^ 2 + (-4 – 1) ^ 2) = √ (4 ^ 2 + 5 ^ 2) = √ (16 + 25) = √41;

| AC | = √ ((2 – 3) ^ 2 + (3 – (-4)) ^ 2) = √ (1 ^ 2 + 7 ^ 2) = √ (1 + 49) = √50.

Since the lengths of all sides of the ABC triangle are different, this triangle is not isosceles.