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 $(5a-3h+6k+100\sin2\alpha)$ is equal to
Solution:
Centroid $G(h,k)$ formula:
$$ h = \frac{6+10\cos\alpha-10\sin\alpha}{3} \Rightarrow 6+10\cos\alpha-10\sin\alpha = 3h \quad \dots(i) $$
$$ k = \frac{8-10\sin\alpha+10\cos\alpha}{3} \Rightarrow 8-10\sin\alpha+10\cos\alpha = 3k \quad \dots(ii) $$
From (ii), $10\cos\alpha-10\sin\alpha = 3k-8$. Substituting into (i):
$$ 6 + (3k-8) = 3h \Rightarrow 3h – 3k + 2 = 0 \quad \dots(iv) $$
Vertices lie on $x^2+y^2=100 \Rightarrow$ Circumcenter $O(0,0)$.
Centroid $G$ divides $LO$ in ratio $2:1$.
$$ G = \left( \frac{1(a)+2(0)}{3}, \frac{1(9)+2(0)}{3} \right) \Rightarrow (h,k) = \left( \frac{a}{3}, 3 \right) $$
So, $k=3$ and $h=a/3$.
Substitute $k=3$ and $h=a/3$ into eq (iv):
$$ 3(a/3) – 3(3) + 2 = 0 \Rightarrow a – 9 + 2 = 0 \Rightarrow a = 7 $$
$$ \therefore h = 7/3 $$
From (ii) with $k=3$:
$$ -10\sin\alpha + 10\cos\alpha = 3(3) – 8 = 1 $$
Squaring both sides:
$$ 100(\sin^2\alpha + \cos^2\alpha – 2\sin\alpha\cos\alpha) = 1 $$
$$ 100(1 – \sin2\alpha) = 1 \Rightarrow 100 – 100\sin2\alpha = 1 $$
$$ 100\sin2\alpha = 99 $$
Value required: $5a – 3h + 6k + 100\sin2\alpha$
$$ = 5(7) – 3(7/3) + 6(3) + 99 $$
$$ = 35 – 7 + 18 + 99 = 145 $$
Ans. 145