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QJEEM24Mac897: JEE Main 2024 Mathematics - Class 11 JEEM24Mac897 Solution: Consider a triangle having the vertices and and angles and . If …

Consider a triangle ABC\mathrm{ABC} having the vertices A(1,2),B(α,β)\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta) and C(γ,δ)\mathrm{C}(\gamma, \delta) and angles ∠ABC=π6\angle A B C=\frac{\pi}{6} and ∠BAC=2π3\angle B A C=\frac{2 \pi}{3} . If the points B\mathrm{B} and C\mathrm{C} lie on the line y=x+4y=x+4 , then α2+γ2\alpha^2+\gamma^2 is equal to _______.

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JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 4 English Explanation 1

amp;P=∣2−1−4∣12+12=32amp;sin⁡(π6)=3/2AB=12⇒AB=62amp;⇒(α−1)2+(α+4−2)2=18\begin{aligned}& P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\& \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{1}{2} \Rightarrow A B=\frac{6}{\sqrt{2}} \\& \Rightarrow(\alpha-1)^2+(\alpha+4-2)^2=18\end{aligned}

⇒2α2+2α−13=0→α\Rightarrow 2 \alpha^2+2 \alpha-13=0 \rightarrow \alpha and γ\gamma satisfy same equation

JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Properties of Triangle Question 4 English Explanation 2

amp;⇒α2+γ2=(α+γ)2−2αγamp;=(−1)2−2(−132)=1+13=14\begin{aligned}& \Rightarrow \alpha^2+\gamma^2=(\alpha+\gamma)^2-2 \alpha \gamma \\& =(-1)^2-2\left(\frac{-13}{2}\right)=1+13=14\end{aligned}

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