Question:medium
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:
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:
Updated On: Jun 6, 2026
- 16
- 27
- 36
- 48
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We are given an equilateral triangle PQR with the coordinates of vertex P and the equation of the line containing the opposite side QR. The task is to find the orthocentre of this triangle.
Step 2: Key Formula or Approach:
1. Special Property of Equilateral Triangles: A critical property is that the orthocentre, centroid, circumcentre, and incentre all coincide at a single point. 2. Use the Centroid: Calculating the centroid's position is typically the most direct method. The centroid divides a median in the ratio 2:1. 3. Altitude and Median Coincidence: In an equilateral triangle, the altitude from a vertex to the opposite side is also the median to that side. Let M be the foot of the perpendicular from P to QR. M is the midpoint of QR. 4. Strategy: We will first find the coordinates of point M by finding the intersection of the line QR and the altitude line PM. Then, we use the section formula to locate the orthocentre (H), which divides PM in the ratio 2:1.
Step 3: Detailed Explanation:
Given vertex P is at $(3, 5)$ and the line for side QR is $x+y-4=0$.
Part 1: Find the foot of the altitude, M
The altitude PM is perpendicular to the line QR.
The slope of line QR ($y = -x+4$) is $m_{QR} = -1$.
The slope of the altitude PM must be the negative reciprocal, so $m_{PM} = 1$.
The equation of the line PM passes through P(3, 5) with slope 1:
$y - 5 = 1(x - 3) \implies y = x+2 \implies x-y = -2$.
Point M is the intersection of lines QR and PM. We solve the system:
1) $x+y = 4$
2) $x-y = -2$
Adding the two equations gives $2x = 2$, so $x=1$.
Substituting $x=1$ into the first equation gives $1+y=4$, so $y=3$.
Thus, the coordinates of M are $(1, 3)$.
Part 2: Find the Orthocentre, H
In an equilateral triangle, the orthocentre H is the same as the centroid. The centroid divides the median PM in the ratio 2:1, with the longer segment being from the vertex P.
Let H have coordinates $(\alpha, \beta)$. Using the section formula:
\[ H(\alpha, \beta) = \left( \frac{1 \cdot x_P + 2 \cdot x_M}{1+2}, \frac{1 \cdot y_P + 2 \cdot y_M}{1+2} \right) \] \[ H(\alpha, \beta) = \left( \frac{1(3) + 2(1)}{3}, \frac{1(5) + 2(3)}{3} \right) \] \[ H(\alpha, \beta) = \left( \frac{3+2}{3}, \frac{5+6}{3} \right) = \left( \frac{5}{3}, \frac{11}{3} \right) \] So, $\alpha = \frac{5}{3}$ and $\beta = \frac{11}{3}$.
Part 3: Evaluate the final expression
We are asked to find $9(\alpha + \beta)$.
\[ \alpha + \beta = \frac{5}{3} + \frac{11}{3} = \frac{16}{3} \] \[ 9(\alpha + \beta) = 9 \times \left(\frac{16}{3}\right) = 3 \times 16 = 48 \] Step 4: Final Answer:
The value of $9(\alpha + \beta)$ is 48.
We are given an equilateral triangle PQR with the coordinates of vertex P and the equation of the line containing the opposite side QR. The task is to find the orthocentre of this triangle.
Step 2: Key Formula or Approach:
1. Special Property of Equilateral Triangles: A critical property is that the orthocentre, centroid, circumcentre, and incentre all coincide at a single point. 2. Use the Centroid: Calculating the centroid's position is typically the most direct method. The centroid divides a median in the ratio 2:1. 3. Altitude and Median Coincidence: In an equilateral triangle, the altitude from a vertex to the opposite side is also the median to that side. Let M be the foot of the perpendicular from P to QR. M is the midpoint of QR. 4. Strategy: We will first find the coordinates of point M by finding the intersection of the line QR and the altitude line PM. Then, we use the section formula to locate the orthocentre (H), which divides PM in the ratio 2:1.
Step 3: Detailed Explanation:
Given vertex P is at $(3, 5)$ and the line for side QR is $x+y-4=0$.
Part 1: Find the foot of the altitude, M
The altitude PM is perpendicular to the line QR.
The slope of line QR ($y = -x+4$) is $m_{QR} = -1$.
The slope of the altitude PM must be the negative reciprocal, so $m_{PM} = 1$.
The equation of the line PM passes through P(3, 5) with slope 1:
$y - 5 = 1(x - 3) \implies y = x+2 \implies x-y = -2$.
Point M is the intersection of lines QR and PM. We solve the system:
1) $x+y = 4$
2) $x-y = -2$
Adding the two equations gives $2x = 2$, so $x=1$.
Substituting $x=1$ into the first equation gives $1+y=4$, so $y=3$.
Thus, the coordinates of M are $(1, 3)$.
Part 2: Find the Orthocentre, H
In an equilateral triangle, the orthocentre H is the same as the centroid. The centroid divides the median PM in the ratio 2:1, with the longer segment being from the vertex P.
Let H have coordinates $(\alpha, \beta)$. Using the section formula:
\[ H(\alpha, \beta) = \left( \frac{1 \cdot x_P + 2 \cdot x_M}{1+2}, \frac{1 \cdot y_P + 2 \cdot y_M}{1+2} \right) \] \[ H(\alpha, \beta) = \left( \frac{1(3) + 2(1)}{3}, \frac{1(5) + 2(3)}{3} \right) \] \[ H(\alpha, \beta) = \left( \frac{3+2}{3}, \frac{5+6}{3} \right) = \left( \frac{5}{3}, \frac{11}{3} \right) \] So, $\alpha = \frac{5}{3}$ and $\beta = \frac{11}{3}$.
Part 3: Evaluate the final expression
We are asked to find $9(\alpha + \beta)$.
\[ \alpha + \beta = \frac{5}{3} + \frac{11}{3} = \frac{16}{3} \] \[ 9(\alpha + \beta) = 9 \times \left(\frac{16}{3}\right) = 3 \times 16 = 48 \] Step 4: Final Answer:
The value of $9(\alpha + \beta)$ is 48.
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