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Find the type of triangle $ABC$ formed whose vertices are $A(1, 0)$, $B(-5, 0)$ and $C(-2, 5)$.
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$A(1,0)$ $B(-5,0)$ $C(-2,5)$
$AB = \sqrt{(-5 - 1)^2 + (0 - 0)^2} = \sqrt{(-6)^2 + 0^2} = \sqrt{36} = 6$
$BC = \sqrt{(-5 + 2)^2 + (0 - 5)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
$CA = \sqrt{(1 + 2)^2 + (0 - 5)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
$\therefore BC = CA$
So, $\triangle ABC$ is isosceles.
$AB = \sqrt{(-5 - 1)^2 + (0 - 0)^2} = \sqrt{(-6)^2 + 0^2} = \sqrt{36} = 6$
$BC = \sqrt{(-5 + 2)^2 + (0 - 5)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
$CA = \sqrt{(1 + 2)^2 + (0 - 5)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
$\therefore BC = CA$
So, $\triangle ABC$ is isosceles.