If P (0,0), Q (0,5) and R (6,0), then Delta P | vedclass

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

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a right angled

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(C) To determine the type of triangle $\Delta PQR$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$1$. Length of $PQ$: $PQ = \sqrt{(0 - 0)^2 + (5 - 0)^2} = \sqrt{0^2 + 5^2} = 5 	ext{ units}$.$2$. Length of $QR$: $QR = \sqrt{(6 - 0)^2 + (0 - 5)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} 	ext{ units}$.$3$. Length of $PR$: $PR = \sqrt{(6 - 0)^2 + (0 - 0)^2} = \sqrt{6^2 + 0^2} = 6 	ext{ units}$.Since the vertices $P(0,0), Q(0,5),$ and $R(6,0)$ lie on the axes,the angle at the origin $P(0,0)$ is $90^\circ$ because the $y$-axis $(PQ)$ is perpendicular to the $x$-axis $(PR)$.Since one angle is $90^\circ$,$\Delta PQR$ is a right-angled triangle.

If $P (0,0), Q (0,5)$ and $R (6,0),$ then $\Delta PQR$ is $\ldots \ldots \ldots$ triangle.

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