If the vertices of Delta PQR are P(3, 10),Q(6 | vedclass

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(C) To determine the type of triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$1$

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an isosceles

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(C) To determine the type of triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$1$. Length of $PQ$: $PQ^2 = (3 - 6)^2 + (10 - 5)^2 = (-3)^2 + (5)^2 = 9 + 25 = 34$. So,$PQ = \sqrt{34}$.$2$. Length of $QR$: $QR^2 = (6 - 1)^2 + (5 - 2)^2 = (5)^2 + (3)^2 = 25 + 9 = 34$. So,$QR = \sqrt{34}$.$3$. Length of $PR$: $PR^2 = (3 - 1)^2 + (10 - 2)^2 = (2)^2 + (8)^2 = 4 + 64 = 68$. So,$PR = \sqrt{68}$.Since $PQ = QR = \sqrt{34}$,two sides of the triangle are equal in length.Therefore,$\Delta PQR$ is an isosceles triangle.

If the vertices of $\Delta PQR$ are $P(3, 10)$,$Q(6, 5)$,and $R(1, 2)$,then $\Delta PQR$ is a $\dots$ triangle.

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