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

Step 1: The coordinates of the vertices are $P(5, 4)$, $Q(-2, 4)$, and $R(a, b)$. The area of the triangle is given as 35 square units. The area of the triangle can be calculated using the formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute the coordinates of $P(5, 4)$, $Q(-2, 4)$, and $R(a, b)$ into the area formula, and set the area equal to 35: $$\frac{1}{2} \left| 5(4 - b) + (-2)(b - 4) + a(4 - 4) \right| = 35$$ Simplifying this equation: $$\left| 20 - 5b - 2b + 8 \right| = 70$$ $$\left| 28 - 7b \right| = 70$$ This gives two cases: 1. $28 - 7b = 70 \quad \Rightarrow \quad b = -6$ 2. $28 - 7b = -70 \quad \Rightarrow \quad b = 14$ Thus, the coordinates of $R$ are $(2, -6)$. Step 2: The centroid $G$ of a triangle is the point where the medians intersect, and its coordinates are the average of the coordinates of the vertices: $$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$$ Substituting the coordinates of $P(5, 4)$, $Q(-2, 4)$, and $R(2, -6)$, we get the centroid as: $$G = \left( \frac{5 + (-2) + 2}{3}, \frac{4 + 4 + (-6)}{3} \right) = \left( \frac{5}{3}, \frac{2}{3} \right)$$ Step 3: The coordinates of the centroid are $(2, \frac{14}{5})$, and using the centroid formula, we calculate the value of $c + 2d$. We have: $$c + 2d = \frac{5}{3} + \frac{4}{3} = 3$$