If the vertices of a triangle are (0, 2),(1, | vedclass

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(D) Let the vertices be $A(0, 2)$,$B(1, 0)$,and $C(3, 1)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(1 - 0)^2

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Isosceles right angled

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(D) Let the vertices be $A(0, 2)$,$B(1, 0)$,and $C(3, 1)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(1 - 0)^2 + (0 - 2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$BC = \sqrt{(3 - 1)^2 + (1 - 0)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$$AC = \sqrt{(3 - 0)^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$Since $AB = BC = \sqrt{5}$,the triangle is isosceles.Also,$(AB)^2 + (BC)^2 = 5 + 5 = 10$ and $(AC)^2 = 10$.Since $(AB)^2 + (BC)^2 = (AC)^2$,the triangle is a right-angled triangle.Therefore,the triangle is an isosceles right-angled triangle.

If the vertices of a triangle are $(0, 2)$,$(1, 0)$,and $(3, 1)$,then the triangle is

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