Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

The length of AB

$= \sqrt{(-3 - 1)^2 + (0 - (-3))^2}\\[1em] = \sqrt{(-4)^2 + 3^2}\\[1em] = \sqrt{16 + 9}\\[1em] = \sqrt{25}\\[1em] = 5$

The length of BC

$= \sqrt{(4 - (-3))^2 + (1 - 0)^2}\\[1em] = \sqrt{(-7)^2 + 1^2}\\[1em] = \sqrt{49 + 1}\\[1em] = \sqrt{50}\\[1em] = 5\sqrt{2}\\[1em]$

The length of CA

$= \sqrt{(4 - 1)^2 + (1 - (-3))^2}\\[1em] = \sqrt{3^2 + 4^2}\\[1em] = \sqrt{9 + 16}\\[1em] = \sqrt{25}\\[1em] = 5$

If ABC is an right angled triangle,

AB² + CA² = $(2\sqrt5)$² = 5² + 5² = 25 + 25 = 50 ⇒ BC²

BC² = AB² + CA² ⇒ the triangle is right angled triangle.

and,

AB = CA ⇒ the triangle is isosceles triangle.

Base of triangle = Height of the triangle = 5 units.

Area of triangle ABC = $\dfrac{1}{2}$ x base x height

$= \dfrac{1}{2} \times 5 \times 5\\[1em] = \dfrac{25}{2}\\[1em] = 12.5 \text{ sq. units}$

Hence, the triangle ABC is an isosceles right-angled triangle and area of the triangle = 12.5 sq. units.