Straight Lines
Straight Lines
nta_pyq_2025_apr
Grade 11
Question:
Let $A(6,8)$, $B(10\cos\alpha,-10\sin\alpha)$ and $C(-10\sin\alpha,10\cos\alpha)$ be the vertices of a triangle. If $L(a,9)$ and $G(h,k)$ be its orthocenter and centroid respectively, then $(5a-3h+6k+100\sin 2\alpha)$ is equal to ____.
Step-by-Step Solution
Key Concept: Use the centroid formula to express $(h,k)$ in terms of $\alpha$, exploit $|AB|^2+|AC|^2-$ orthocenter condition to derive $\sin 2\alpha$, find $a$ from $k=9/3$ and solve for each required quantity.
Centroid: $k = \dfrac{8-10\sin\alpha+10\cos\alpha}{3}$. Also $h = \dfrac{6+10\cos\alpha-10\sin\alpha}{3}$.
From orthocenter $L(a,9)$: using $k=3$ (from equations) and $a=3h$:
$10(\cos\alpha-\sin\alpha) = 1$, $\Rightarrow 100\sin 2\alpha = 99$.
$h = 7/3$, $a=7$, $k=3$.
$$5a-3h+6k+100\sin 2\alpha = 35-7+18+99 = 145.$$
Correct Answer: 145