If P, Q, R are collinear points such that P(7 | vedclass

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(A) Given $P(7, 7)$ and $Q(3, 4)$.First,calculate the distance $PQ = \sqrt{(7-3)^2 + (7-4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5$.Since $P, Q, R$ ar

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$(-1, 1)$

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(A) Given $P(7, 7)$ and $Q(3, 4)$.First,calculate the distance $PQ = \sqrt{(7-3)^2 + (7-4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5$.Since $P, Q, R$ are collinear and $PR = 10$,and $PQ = 5$,$Q$ is the midpoint of $PR$.Let $R = (h, k)$.Using the midpoint formula: $\frac{7+h}{2} = 3$ $\Rightarrow 7+h = 6$ $\Rightarrow h = -1$.$\frac{7+k}{2} = 4$ $\Rightarrow 7+k = 8$ $\Rightarrow k = 1$.Therefore,the coordinates of $R$ are $(-1, 1)$.

If $P, Q, R$ are collinear points such that $P(7, 7)$,$Q(3, 4)$ and $PR = 10$,then what are the coordinates of $R$?

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