In an equilateral triangle PQR, let the verte | vedclass

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(D) In an equilateral triangle, the orthocentre coincides with the centroid.Let $P = (3, 5)$ and the line $QR$ be $x + y - 4 = 0$.The altitude from $P

Vedclass · Accepted Answer

$48$

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(D) In an equilateral triangle, the orthocentre coincides with the centroid.Let $P = (3, 5)$ and the line $QR$ be $x + y - 4 = 0$.The altitude from $P$ to $QR$ is perpendicular to $x + y = 4$. The slope of $QR$ is $-1$, so the slope of the altitude is $1$.The equation of the altitude passing through $(3, 5)$ is $y - 5 = 1(x - 3)$, which simplifies to $y = x + 2$.The foot of the altitude $F$ is the intersection of $x + y = 4$ and $y = x + 2$. Substituting $y$, we get $x + (x + 2) = 4$, so $2x = 2$, $x = 1$. Then $y = 3$. Thus, $F = (1, 3)$.The centroid $G(\alpha, \beta)$ divides the altitude $PF$ in the ratio $2:1$ from the vertex $P$.Using the section formula, $G = \left( \frac{2(1) + 1(3)}{2 + 1}, \frac{2(3) + 1(5)}{2 + 1} \right) = \left( \frac{5}{3}, \frac{11}{3} \right)$.Thus, $\alpha = 5/3$ and $\beta = 11/3$.Therefore, $9(\alpha + \beta) = 9(5/3 + 11/3) = 9(16/3) = 48$.

In an equilateral triangle $PQR$, let the vertex $P$ be at $(3, 5)$ and the side $QR$ be along the line $x + y = 4$. If the orthocentre of the triangle $PQR$ is $(\alpha, \beta)$, then $9(\alpha + \beta)$ is equal to:

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