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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Straight Lines

Let A(6,8), B(10 α,-10 α) and C(-10 α, 10 α) , be the vertices of a...

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

Correct answer

0

Step-by-step solution

a+03=h⇒a=3h9+03=k⇒k=3∵(h,k)=(6+10cos⁡α−10sin⁡α3,8−10sin⁡α−10cos⁡α3)6+10cos⁡α−10sin⁡α=3h10cos⁡α−10sin⁡α=3h−610(cos⁡α−sin⁡α)=18−10sin⁡α+10cos⁡α3=k⇒100sin⁡2α=99h=73⇒a=7\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$
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