Let A(6, 8), B(10 cos , – 10 sin ) and C(–10 sin , 10 cos ), 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 in 2) is equal to __________. [2025]

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Solution

Q.
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 (5a – 3h + 6k + 100 sin 2 $α$ ) is equal to __________. [2025]

Ans.
(145)

We can observe that all the three points A, B, C lie on the circle $x^{2} + y^{2} = 100$ , so circumcentre is (0, 0). Since, centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1, then

$\frac{a + 0}{3} = h \Rightarrow a = 3 h$

and $\frac{9 + 0}{3} = k \Rightarrow k = 3$

Also, $\frac{6 + 10 \cos α - 10 \sin α}{3} = h$

$\Rightarrow 10 (\cos α - \sin α) = 3 h - 6$ ... (i)

and $\frac{8 + 10 \cos α - 10 \sin α}{3} = k$

$\Rightarrow 10 (\cos α - \sin α) = 3 k - 8 = 9 - 8 = 1$ ... (ii)

On squaring, 100 (1 – sin 2 $α$ ) = 1 $\Rightarrow$ 100 sin 2 $α$ = 99

From (i) and (ii), we get $h = \frac{7}{3}$

Now, 5a – 3h + 6k +100 sin 2 $α$

= 15h – 3h + 6k + 100 sin 2 $α$ = $12 \times \frac{7}{3} + 18 + 99$ = 145.

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