The area of a $\triangle PQR$ with vertices $P(5,4)$, $Q(-2,4)$ and $R(a,b)$ be $35$ square units. If its orthocenter and centroid are $O\left(\frac{14}{5},2\right)$ and $C(c,d)$ respectively, then $c+2d$ is equal to
- (1) $\dfrac{7}{3}$
- (2) $3$
- (3) $2$
- (4) $\dfrac{8}{3}$
Solution:
Points $P(5,4)$ and $Q(-2,4)$ have the same y–coordinate, hence $PQ$ is a horizontal line.
For a triangle with one horizontal side, the altitude from the opposite vertex is vertical.
Therefore, the x–coordinate of vertex $R$ is equal to the x–coordinate of the orthocenter.
$$R\left(\frac{14}{5},\,b\right)$$
Length of base:
$$PQ = |5 – (-2)| = 7$$
Height of triangle:
$$|b – 4|$$
Using area formula,
$$35 = \frac{1}{2} \times 7 \times |b – 4|$$
$$|b – 4| = 10$$
$$b = 14 \text{ or } -6$$
Since the orthocenter lies inside the triangle, the triangle must lie below $PQ$.
Hence,
$$R(2,-6)$$
Centroid of triangle:
$$C\left(\frac{5 + (-2) + 2}{3},\,\frac{4 + 4 + (-6)}{3}\right)
= \left(\frac{5}{3},\,\frac{2}{3}\right)$$
$$c + 2d = \frac{5}{3} + 2\left(\frac{2}{3}\right) = 3$$
Ans. (2)