Let $\mathrm{A}(6,8), \quad \mathrm{B}(10 \quad \cos \alpha,-10 \quad \sin \alpha)$ and $\mathrm{C}(-10 \sin \alpha, 10 \cos \alpha)$, be the vertices of a triangle. If $\mathrm{L}(\mathrm{a}, 9)$ and $\mathrm{G}(\mathrm{h}, \mathrm{k})$ be its orthocenter and centroid respectively, then $(5 a-3 h+6 k+100 \sin 2 \alpha)$ is equal to ____

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Straight Lines

Let

A (6, 8), B (10 cos α, - 10 sin α)

and

C (- 10 sin α, 10 cos α)

, 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 α)

is equal to ____

Answer: 145

Note

∣ A ∣ = 36 + 64 = 10

, and

∣ B ∣ = ∣ C ∣ = 100 cos^{2} α + 100 sin^{2} α = 10

, so A, B, C all lie on the circle

x^{2} + y^{2} = 100

; hence the circumcenter is

O = (0, 0)

.

By the Euler line property,

O

, the centroid

G

, and the orthocenter

L

are collinear with

L = O + 3 (G - O) = 3 G

(since

O

is the origin). So

(a, 9) = 3 (h, k) \Rightarrow a = 3 h, k = 3.

The centroid's y-coordinate:

k = \frac{8 + ( - 10 sin α ) + 10 cos α}{3} = 3 \Rightarrow 10 (cos α - sin α) = 1.

Squaring:

1 - sin 2 α = \frac{1}{100} \Rightarrow 100 sin 2 α = 99

.

The centroid's x-coordinate:

h = \frac{6 + 10 cos α - 10 sin α}{3} = \frac{6 + 1}{3} = \frac{7}{3}, a = 3 h = 7.

5 a - 3 h + 6 k + 100 sin 2 α = 5 (7) - 3 (\frac{7}{3}) + 6 (3) + 99 = 35 - 7 + 18 + 99 = 145.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Integer answer