Show that A(0, 0), B(5, 5) and C(-5, 5) are vertices of a right angled isosceles triangle.

Distance between 2 points (x₁, y₁) and (x₂, y₂) = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Distance between A(0, 0) and B(5, 5) =

$= \sqrt{(5 - 0)^2 + (5 - 0)^2}\\[1em] = \sqrt{5^2 + 5^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}$

Distance between B(5, 5) and C(-5, 5) =

$= \sqrt{(-5 - 5)^2 + (5 - 5)^2}\\[1em] = \sqrt{(-10)^2}\\[1em] = \sqrt{100}\\[1em] = 10$

Distance between A(0, 0) and C(-5, 5) =

$= \sqrt{(-5 - 0)^2 + (5 - 0)^2}\\[1em] = \sqrt{(-5)^2 + 5^2}\\[1em] = \sqrt{25 + 25}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}$

Using pythagoras theorem,

BC² = AB² + AC²

$\Rightarrow (\sqrt{100})^2 = (5\sqrt{2})^2 + (5\sqrt{2})^2$

⇒ 100 = 50 + 50

⇒ 100 = 100

BC² = AB² + AC² ⇒ the triangle is right-angled triangle.

And, AB = AC ⇒ the triangle is an isosceles.

Hence, the triangle ABC is an isosceles right-angled triangle.