Question

Let the area of a triangle \( \triangle PQR \) with vertices \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \) be 35 square units. If its orthocenter and centroid are \( O(2, \frac{14}{5}) \) and \( C(c, d) \) respectively, then \( c + 2d \) is equal to:

• A. \( \frac{7}{3} \)
• B. \( 3 \)
• C. \( 2 \)
• D. \( \frac{8}{3} \)

Hint:

The centroid of a triangle can be calculated as the average of the coordinates of the three vertices. Additionally, the area of a triangle can be used to find relationships between the coordinates of the points.

Answer 1

The Correct Option is B

To solve the problem, we need to find the value of \( c + 2d \) where \( (c, d) \) is the centroid of triangle \( PQR \) with given points \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \), and orthocenter \( O(2, \frac{14}{5}) \).

1. Calculating the Area of Triangle \( PQR \):
The area of triangle \( PQR \) is given by:

\[ \text{Area} = \frac{1}{2} |5(4-b) - 2(b-4) + a(4-4)| = \frac{1}{2} |20 - 5b - 2b + 8| = \frac{1}{2} |28 - 7b| = \frac{7}{2}|4-b| \]
Given that the area is 35, we have:

\[ \frac{7}{2}|4-b| = 35 \implies |4-b| = 10 \]
This gives two cases:

\[ 4 - b = 10 \implies b = -6 \quad \text{or} \quad 4 - b = -10 \implies b = 14 \]

2. Finding the Centroid \( C(c, d) \):
The centroid coordinates are:

\[ c = \frac{5 + (-2) + a}{3} = \frac{3 + a}{3} \quad \text{and} \quad d = \frac{4 + 4 + b}{3} = \frac{8 + b}{3} \]

3. Case 1: \( b = -6 \):
Here, \( R(a, -6) \), and \( d = \frac{8 - 6}{3} = \frac{2}{3} \).
Since \( PQ \) is horizontal (slope \( 0 \)), the altitude from \( R \) is vertical, so its equation is \( x = a \).
Given the orthocenter \( O(2, \frac{14}{5}) \), we have \( a = 2 \).
Thus, \( c = \frac{3 + 2}{3} = \frac{5}{3} \).
Now, \( c + 2d = \frac{5}{3} + 2 \left( \frac{2}{3} \right) = \frac{5 + 4}{3} = \frac{9}{3} = 3 \).

4. Case 2: \( b = 14 \):
Here, \( R(a, 14) \), and \( d = \frac{8 + 14}{3} = \frac{22}{3} \).
Again, the altitude from \( R \) is vertical, so \( a = 2 \).
Thus, \( c = \frac{5}{3} \), and \( c + 2d = \frac{5}{3} + 2 \left( \frac{22}{3} \right) = \frac{49}{3} \).
However, verifying the orthocenter condition leads to a contradiction, so this case is invalid.

5. Conclusion:
Only \( b = -6 \) is valid, giving \( R(2, -6) \), \( d = \frac{2}{3} \), and \( c = \frac{5}{3} \).
Thus, \( c + 2d = 3 \).

Final Answer:
The final answer is \(\boxed{3}\).