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Show that the points $(-2, 3)$, $(8, 3)$ and $(6, 7)$ are the vertices of a right-angled triangle.
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Let the given points be A $(-2, 3)$, B $(8, 3)$ and C $(6, 7)$
Then, AB = $10$, BC = $\sqrt{4 + 16} = \sqrt{20}$,
AC = $\sqrt{64 + 16} = \sqrt{80}$
$\therefore AB^2 = BC^2 + AC^2$
$\therefore$ the given points are the vertices of a right angled triangle.
Then, AB = $10$, BC = $\sqrt{4 + 16} = \sqrt{20}$,
AC = $\sqrt{64 + 16} = \sqrt{80}$
$\therefore AB^2 = BC^2 + AC^2$
$\therefore$ the given points are the vertices of a right angled triangle.