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Prove that the points $(3, 0), (6, 4)$ and $(-1, 3)$ are the vertices of an isosceles triangle.
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Let $A(3,0), B(6,4), C(-1, 3)$
$AB = \sqrt{(3 - 6)^2 + (0 - 4)^2} = 5$
$BC = \sqrt{(6 + 1)^2 + (4 - 3)^2} = \sqrt{50}$
$CA = \sqrt{(3 + 1)^2 + (0 - 3)^2} = 5$
As, $AB = AC$
$\therefore ABC$ is an isosceles triangle
$AB = \sqrt{(3 - 6)^2 + (0 - 4)^2} = 5$
$BC = \sqrt{(6 + 1)^2 + (4 - 3)^2} = \sqrt{50}$
$CA = \sqrt{(3 + 1)^2 + (0 - 3)^2} = 5$
As, $AB = AC$
$\therefore ABC$ is an isosceles triangle