Question

Consider an equilateral triangle PQR, where P(3, 5) and QR is \( x + y = 4 \). The orthocentre of \( \Delta PQR \) is \( (\alpha, \beta) \), then \( 9(\alpha + \beta) \) is equal to:

• A. 56
• B. 48
• C. 64
• D. 36

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In an equilateral triangle, the orthocentre, circumcentre, incentre, and centroid all coincide at the same point. The centroid divides the median (altitude in this case) in the ratio 2:1 from the vertex.
Step 2: Key Formula or Approach:
1. Find the foot of the perpendicular (\( M \)) from \( P(3, 5) \) to the line \( x + y - 4 = 0 \). 2. The orthocentre (\( O \)) divides the segment \( PM \) in the ratio 2:1.
Step 3: Detailed Explanation:
1. Foot of perpendicular \( M(x, y) \): \[ \frac{x - 3}{1} = \frac{y - 5}{1} = -\frac{(3 + 5 - 4)}{1^2 + 1^2} = -\frac{4}{2} = -2 \] \[ x = 3 - 2 = 1, \quad y = 5 - 2 = 3 \implies M(1, 3) \] 2. Section formula for Orthocentre \( (\alpha, \beta) \) dividing \( PM \) in ratio 2:1: \[ \alpha = \frac{2(1) + 1(3)}{2+1} = \frac{5}{3} \] \[ \beta = \frac{2(3) + 1(5)}{2+1} = \frac{11}{3} \] 3. Find \( 9(\alpha + \beta) \): \[ \alpha + \beta = \frac{5}{3} + \frac{11}{3} = \frac{16}{3} \] \[ 9(\alpha + \beta) = 9 \times \frac{16}{3} = 3 \times 16 = 48 \]
Step 4: Final Answer:
The value of \( 9(\alpha + \beta) \) is 48.