Question

Prove that the points (2, −2), (−2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the area of this triangleb ?

Answer 1

Let A(2, −2), B(−2, 1) and C(5, 2) be the vertices of the given triangle.

Now,

AB =\[\sqrt{\left( 1 + 2 \right)^2 + \left( - 2 - 2 \right)^2} = \sqrt{25} = 5 units\]

BC =\[\sqrt{\left( 1 + 2 \right)^2 + \left( - 2 - 2 \right)^2} = \sqrt{25} = 5 units\]

CA =\[\sqrt{\left( 2 + 2 \right)^2 + \left( 5 - 2 \right)^2} = \sqrt{25} = 5 units\]

∴ CA²  + AB² = BC²

Hence, by the converse of Pythagoras' theorem, ∠A = 90°.

∆ABC is right-angled.