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 , then is equal to: [2026]

Question

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 $(α, β)$ , then $9 (α + β)$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

AP is perpendicular to QR

FIGURE

Slope of QR = –1

Equation of AP is y – 5 = 1(x – 3)

$\Rightarrow y - 5 = x - 3$

$\Rightarrow x - y = - 2$

On solving, x + y = 4 and x – y = –2, we get

x = 1 and y = 3

$∴ A \equiv (1, 3)$

Now, $G \equiv (\frac{2 + 3}{3}, \frac{6 + 5}{3}) = (5 / 3, 11 / 3) = (α, β)$

$∴ 9 (α + β) = 9 (\frac{5}{3} + \frac{11}{3}) = 9 \times \frac{16}{3} = 48$

Solution

Q.
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 $(α, β)$ , then $9 (α + β)$ is equal to: [2026]

1 16

2 27

3 36

4 48

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