Question

Prove that the points \( (3, 2) \), \( (-2, -3) \), and \( (2, 3) \) are vertices of a right angled triangle.

Hint:

To prove that three points form a right angled triangle, calculate the distances between the points and check if the Pythagorean theorem holds.

Answer 1

Let the three points be \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \). We need to prove that the triangle formed by these points is a right angled triangle. To do this, we will use the distance formula to calculate the lengths of the sides of the triangle and check if the Pythagorean theorem holds. The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \] Step 1: Find the lengths of the sides. - Distance \( AB \): \[ AB = \sqrt{(3 - (-2))^2 + (2 - (-3))^2} = \sqrt{(3 + 2)^2 + (2 + 3)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}. \] - Distance \( BC \): \[ BC = \sqrt{(2 - (-2))^2 + (3 - (-3))^2} = \sqrt{(2 + 2)^2 + (3 + 3)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}. \] - Distance \( CA \): \[ CA = \sqrt{(3 - 2)^2 + (2 - 3)^2} = \sqrt{(3 - 2)^2 + (2 - 3)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}. \] Step 2: Check the Pythagorean theorem. For a right angled triangle, the square of the hypotenuse should be equal to the sum of the squares of the other two sides. We will check if the squares of \( AB \), \( BC \), and \( CA \) satisfy this relation. \[ AB^2 + CA^2 = (5\sqrt{2})^2 + (\sqrt{2})^2 = 50 + 2 = 52. \] \[ BC^2 = (2\sqrt{13})^2 = 52. \] Since \( AB^2 + CA^2 = BC^2 \), the points \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \) form a right angled triangle.