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 ($\alpha, \beta$), then 9($\alpha + \beta$) is equal to:

• A. 16
• B. 27
• C. 36
• D. 48

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an equilateral triangle with one vertex P and the equation of the line containing the opposite side QR. We need to find the coordinates of the orthocentre and then evaluate a given expression.
Step 2: Key Formula or Approach:
1. A key property of an equilateral triangle is that its orthocentre, centroid, circumcentre, and incentre all coincide at the same point.
2. It's easiest to work with the centroid. The centroid divides the median in the ratio 2:1 from the vertex to the midpoint of the opposite side.
3. In this case, the altitude from P to QR is also the median. Let M be the foot of the perpendicular from P to QR. M is the midpoint of QR.
4. The orthocentre (which is also the centroid) H lies on the altitude PM and divides it such that PH:HM = 2:1.
5. We will find the coordinates of M (the foot of the perpendicular from P to the line QR) and then use the section formula to find the coordinates of H.
Step 3: Detailed Explanation:
Vertex P is at $(3, 5)$.
The line containing side QR is $L: x+y-4=0$.
Find the foot of the altitude (M):
Let M be the foot of the perpendicular from P(3, 5) to the line QR. The altitude PM is a line perpendicular to QR and passing through P.
The slope of line QR ($y = -x+4$) is $m_{QR} = -1$.
The slope of the altitude PM will be $m_{PM} = -\frac{1}{m_{QR}} = -\frac{1}{-1} = 1$.
The equation of the line PM passing through P(3, 5) with slope 1 is:
$y - 5 = 1(x - 3) \implies y = x+2 \implies x-y+2=0$.
The point M is the intersection of lines QR and PM.
QR: $x+y=4$ --- (1)
PM: $x-y=-2$ --- (2)
Adding (1) and (2):
$2x = 2 \implies x=1$.
Substituting $x=1$ into (1):
$1+y=4 \implies y=3$.
So, the coordinates of M are $(1, 3)$.
Find the Orthocentre (H):
The orthocentre H coincides with the centroid. The centroid divides the median PM in the ratio 2:1. The coordinates of H($\alpha, \beta$) can be found using the section formula:
$H = \left( \frac{2 \cdot x_M + 1 \cdot x_P}{2+1}, \frac{2 \cdot y_M + 1 \cdot y_P}{2+1} \right)$
$H(\alpha, \beta) = \left( \frac{2(1) + 1(3)}{3}, \frac{2(3) + 1(5)}{3} \right)$
$H(\alpha, \beta) = \left( \frac{2+3}{3}, \frac{6+5}{3} \right)$
$H(\alpha, \beta) = \left( \frac{5}{3}, \frac{11}{3} \right)$.
So, $\alpha = \frac{5}{3}$ and $\beta = \frac{11}{3}$.
Evaluate the expression:
We need to find the value of $9(\alpha + \beta)$.
$\alpha + \beta = \frac{5}{3} + \frac{11}{3} = \frac{16}{3}$.
$9(\alpha + \beta) = 9 \times \frac{16}{3} = 3 \times 16 = 48$.
Step 4: Final Answer:
The value of $9(\alpha + \beta)$ is 48.