Straight Lines
Straight Lines
nta_pyq_2025_apr
Grade 11
Question:
Let 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 $O\!\left(2,\dfrac{14}{5}\right)$ and $C(c,d)$ respectively, then $c+2d$ is equal to
$\dfrac{8}{3}$
$\dfrac{7}{3}$
$2$
$3$
Step-by-Step Solution
Key Concept: Use the orthocenter conditions (altitude from $P$ $\perp$ $QR$, altitude from $Q$ $\perp$ $PR$) to find $R(a,b)$, then compute the centroid and evaluate $c+2d$.
Altitude from $P(5,4)$ perpendicular to $QR$, and $O(2,14/5)$ lies on this altitude: gives line $QR: 5x+2y+2=0$.
Altitude from $Q(-2,4)$ perpendicular to $PR$: gives line $PR: 10x-3y-38=0$.
Intersecting: $R(2,-6)$.
Centroid $C = \left(\dfrac{5-2+2}{3}, \dfrac{4+4-6}{3}\right) = \left(\dfrac{5}{3},\dfrac{2}{3}\right)$.
$$c+2d = \frac{5}{3}+\frac{4}{3} = 3.$$
Correct Answer: 4