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

Let us prove that the triangle with vertices A (2; 3), B (-1; -1) and C (3; -4) is isosceles.

Find the sides of the triangle.

Use the formula √ ((x2 – x1) ^ 2 + (y2 – y1) ^ 2) to find the sides of a triangle.

AB = √ ((- 1 – 2) ^ 2 + (-1 – 3) ^ 2) = √ ((- 3) ^ 2 + (-4) ^ 2) = √ (9 + 16) = √25 = √5 ^ 2 = 5;

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

BC = √ ((3 – (-1)) ^ 2 + (-4 – (-1)) ^ 2) = √ ((3 + 1) ^ 2 + (-4 + 1) ^ 2) = √ ( 4 ^ 2 + (-3) ^ 2) = √ (16 + 9) = √25 = 5;

Since the sides AB and BC are equal, it means that the triangle is isosceles.