JEE Main 2024 (Online) 4th April Evening ShiftQuestion 1
Consider a triangle $ABC$ having the vertices $A(1,2)$, $B(\alpha,\beta)$ and $C(\gamma,\delta)$ and angles $\angle ABC=\dfrac{\pi}{6}$ and $\angle BAC=\dfrac{2\pi}{3}$. If the points $B$ and $C$ lie on the line $y=x+4$, then $\alpha^{2}+\gamma^{2}$ is equal to $\_\_\_\_\_\_\_\_$ .
Points and Straight Lines
Answer$14$
Solution
$\angle C=180^\circ-120^\circ-30^\circ=30^\circ=\angle B$, so $AB=AC$ and the foot $M$ of the perpendicular from $A$ to $y=x+4$ is the midpoint of $BC$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 225"><path stroke-width=".8" d="M12 193.1h186m-88.1 19.6V12"/><path fill="#000" stroke="none" d="m198 193.1-5-2.2v4.4zM109.9 12l-2.2 5h4.4z"/><text x="203.6" y="202.9" fill="#000" stroke="none" font-style="italic">x</text><text x="115.9" y="15.5" fill="#000" stroke="none" font-style="italic">y</text><text x="104.3" y="202.9" fill="#000" stroke="none" font-style="italic" text-anchor="end">O</text><path d="M12 193.1 193.1 12"/><text x="187.4" y="28.2" fill="#000" stroke="none" font-size="8.5">y=x+4</text><path stroke-width="1.1" d="M134.4 144.2 34.1 171 161.2 43.9Z"/><path stroke-dasharray="4 3" d="m134.4 144.2-36.7-36.8"/><path stroke-width=".8" d="m101.9 111.7 4.2-4.2-4.2-4.2"/><circle cx="134.4" cy="144.2" r="1.9" fill="#000" stroke-width=".7"/><text x="141.4" y="147.7" fill="#000" stroke="none">A(1,2)</text><circle cx="34.1" cy="171" r="1.9" fill="#000" stroke-width=".7"/><text x="28.5" y="168.9" fill="#000" stroke="none" text-anchor="end">B</text><circle cx="161.2" cy="43.9" r="1.9" fill="#000" stroke-width=".7"/><text x="155.6" y="41.8" fill="#000" stroke="none" text-anchor="end">C</text><circle cx="97.7" cy="107.4" r="1.9" fill="#000" stroke-width=".7"/><text x="92.1" y="105.3" fill="#000" stroke="none" text-anchor="end">M</text><path stroke-width=".8" d="M123.7 147a11 11 0 0 1 13.5-13.5"/><text x="142.1" y="155.4" fill="#000" stroke="none" font-size="8.5" text-anchor="middle">120°</text><path stroke-width=".8" d="M51 154.1a24 24 0 0 1 6.3 10.7"/><text x="61.8" y="158.5" fill="#000" stroke="none" font-size="8.5" text-anchor="middle">30°</text></svg> $AM=\frac{|1-2+4|}{\sqrt2}=\frac{3}{\sqrt2}$ and $M=\left(-\frac12,\frac72\right)$. In $\triangle ABM$: $MB=AM\cot30^\circ=\frac{3\sqrt3}{\sqrt2}$. Along the line, unit direction $\frac{1}{\sqrt2}(1,1)$: $B,C=M\pm\frac{3\sqrt3}{2}(1,1)$, so $\alpha,\gamma=-\frac12\pm\frac{3\sqrt3}{2}$. $\alpha^2+\gamma^2=2\cdot\frac14+2\cdot\frac{27}{4}=14$.