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 $(5 a-3 h+6 k+100 \sin 2 \alpha)$ is equal to ______ -.

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \frac{a+0}{3}=h \Rightarrow a=3 h \\ & \frac{9+0}{3}=k \Rightarrow k=3 \\ & \because(h, k)=\left(\frac{6+10 \cos \alpha-10 \sin \alpha}{3}, \frac{8-10 \sin \alpha-10 \cos \alpha}{3}\right) \\ & 6+10 \cos \alpha-10 \sin \alpha=3 h \\ & 10 \cos \alpha-10 \sin \alpha=3 h-6 \\ & 10(\cos \alpha-\sin \alpha)=1 \\ & \frac{8-10 \sin \alpha+10 \cos \alpha}{3}=k \\ & \Rightarrow 100 \sin 2 \alpha=99 \\ & h=\frac{7}{3} \\ & \Rightarrow a=7
\end{aligned}$
$\text { Now, } 5 a-3 h+6 k+100 \sin 2 \alpha=35-7+18+99$
$=145$

Let $A(6,8), B(10 \cos \alpha,-10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$, be the vertices of…

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