JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 24)
Consider a triangle $$\mathrm{ABC}$$ having the vertices $$\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$$ and $$\mathrm{C}(\gamma, \delta)$$ and angles $$\angle A B C=\frac{\pi}{6}$$ and $$\angle B A C=\frac{2 \pi}{3}$$. If the points $$\mathrm{B}$$ and $$\mathrm{C}$$ lie on the line $$y=x+4$$, then $$\alpha^2+\gamma^2$$ is equal to _______.
Answer
14
Explanation
_4th_April_Evening_Shift_en_24_1.png)
$$\begin{aligned} & P=\frac{|2-1-4|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}} \\ & \sin \left(\frac{\pi}{6}\right)=\frac{3 / \sqrt{2}}{A B}=\frac{1}{2} \Rightarrow A B=\frac{6}{\sqrt{2}} \\ & \Rightarrow(\alpha-1)^2+(\alpha+4-2)^2=18 \end{aligned}$$
$$\Rightarrow 2 \alpha^2+2 \alpha-13=0 \rightarrow \alpha$$ and $$\gamma$$ satisfy same equation
_4th_April_Evening_Shift_en_24_2.png)
$$\begin{aligned} & \Rightarrow \alpha^2+\gamma^2=(\alpha+\gamma)^2-2 \alpha \gamma \\ & =(-1)^2-2\left(\frac{-13}{2}\right)=1+13=14 \end{aligned}$$
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