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Using distance formula, show that the points $(-1, 3), (6, 2)$ and $(3, -1)$ are vertices of a right-angled triangle.
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Solution: Let $A(-1, 3), B(6, 2)$ and $C(3, -1)$ are given points
$AB = \sqrt{(6 + 1)^2 + (2 - 3)^2} = \sqrt{50}$
$BC = \sqrt{(3 - 6)^2 + (-1 - 2)^2} = \sqrt{18}$
$CA = \sqrt{(-1 - 3)^2 + (3 + 1)^2} = \sqrt{32}$
Since $AB^2 = BC^2 + CA^2$
$\Rightarrow \Delta ABC$ is right angled at C
$AB = \sqrt{(6 + 1)^2 + (2 - 3)^2} = \sqrt{50}$
$BC = \sqrt{(3 - 6)^2 + (-1 - 2)^2} = \sqrt{18}$
$CA = \sqrt{(-1 - 3)^2 + (3 + 1)^2} = \sqrt{32}$
Since $AB^2 = BC^2 + CA^2$
$\Rightarrow \Delta ABC$ is right angled at C