Let the area of a triangle PQR with vertices | vedclass

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(B) The altitude from $P$ to $QR$ is a vertical line since $P$ and $Q$ have the same $y$-coordinate $(y=4)$. The line $PQ$ is horizontal. The altitude

Vedclass · Accepted Answer

$3$

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(B) The altitude from $P$ to $QR$ is a vertical line since $P$ and $Q$ have the same $y$-coordinate $(y=4)$. The line $PQ$ is horizontal. The altitude from $P$ to $QR$ passes through $P(5, 4)$ and the orthocenter $O\left(2, \frac{14}{5}\right)$. Since $QR$ is perpendicular to $PO$,the slope of $PO$ is $m_{PO} = \frac{4 - 14/5}{5 - 2} = \frac{6/5}{3} = \frac{2}{5}$. Thus,the slope of $QR$ is $m_{QR} = -\frac{5}{2}$.Since $Q(-2, 4)$ lies on $QR$,the equation of $QR$ is $y - 4 = -\frac{5}{2}(x + 2) \implies 2y - 8 = -5x - 10 \implies 5x + 2y + 2 = 0$.The altitude from $Q$ to $PR$ passes through $Q(-2, 4)$ and $O\left(2, \frac{14}{5}\right)$. The slope of $QO$ is $m_{QO} = \frac{14/5 - 4}{2 - (-2)} = \frac{-6/5}{4} = -\frac{3}{10}$. Thus,the slope of $PR$ is $m_{PR} = \frac{10}{3}$.The equation of $PR$ is $y - 4 = \frac{10}{3}(x - 5) \implies 3y - 12 = 10x - 50 \implies 10x - 3y - 38 = 0$.Solving $5x + 2y = -2$ and $10x - 3y = 38$: Multiply the first by $2$: $10x + 4y = -4$. Subtracting: $7y = -42 \implies y = -6$. Then $5x + 2(-6) = -2 \implies 5x = 10 \implies x = 2$. So $R(2, -6)$.The area is $\frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)| = 35 \implies \frac{1}{2} |5(4 - (-6)) + (-2)(-6 - 4) + 2(4 - 4)| = 35 \implies \frac{1}{2} |50 + 20 + 0| = 35$,which is consistent.The centroid $C(c, d) = \left(\frac{5 - 2 + 2}{3}, \frac{4 + 4 - 6}{3}\right) = \left(\frac{5}{3}, \frac{2}{3}\right)$.Therefore,$c + 2d = \frac{5}{3} + 2\left(\frac{2}{3}\right) = \frac{5+4}{3} = 3$.

Let the area of a $\triangle PQR$ with vertices $P(5, 4)$,$Q(-2, 4)$,and $R(a, b)$ be $35$ square units. If its orthocenter and centroid are $O\left(2, \frac{14}{5}\right)$ and $C(c, d)$ respectively,then $c+2d$ is equal to:

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