JEE Main 2024 (Online) 29th January Morning ShiftQuestion 1
In a $\triangle ABC$, suppose $y=x$ is the equation of the bisector of the angle $B$ and the equation of the side $AC$ is $2x-y=2$. If $2AB=BC$ and the points $A$ and $B$ are respectively $(4,6)$ and $(\alpha,\beta)$, then $\alpha+2\beta$ is equal to
Points and Straight Lines
Solution
Let the bisector $y=x$ meet $AC$ at $D$: $2x-x=2\Rightarrow D=(2,2)$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 248"><path stroke-dasharray="4 3" d="M12 198 198 12"/><text x="26.4" y="194.1" fill="#000" stroke="none" font-size="8.5">y=x</text><path stroke-width="1.1" d="m83.5 107.4 95.4-76.3L26.3 221.8Z"/><circle cx="83.5" cy="107.4" r="1.9" fill="#000" stroke-width=".7"/><text x="76.5" y="110.9" fill="#000" stroke="none" text-anchor="end">A(4,6)</text><circle cx="178.9" cy="31.1" r="1.9" fill="#000" stroke-width=".7"/><text x="185.9" y="34.6" fill="#000" stroke="none">B</text><circle cx="26.3" cy="221.8" r="1.9" fill="#000" stroke-width=".7"/><text x="33.3" y="225.3" fill="#000" stroke="none">C</text><circle cx="64.5" cy="145.5" r="1.9" fill="#000" stroke-width=".7"/><text x="71.5" y="149" fill="#000" stroke="none">D(2,2)</text><circle cx="102.6" cy="126.5" r="1.9" fill="#000" stroke-width=".7"/><text x="109.6" y="130" fill="#000" stroke="none">A'(6,4)</text><text x="63.8" y="166.2" fill="#000" stroke="none" font-size="8.5">AD:DC=1:2</text></svg> Angle-bisector theorem: $\frac{AD}{DC}=\frac{AB}{BC}=\frac12$, so $D=\frac{2A+C}{3}$ and $C=3D-2A=(-2,-6)$ (it lies on $2x-y=2$). The image of $A$ in the bisector, $A'=(6,4)$, lies on $BC$. Line $A'C$: slope $\frac{10}{8}=\frac54$, $y-4=\frac54(x-6)$. With $y=x$: $x-4=\frac54x-\frac{15}{2}\Rightarrow x=14$, so $B=(14,14)$ (check: $AB=\sqrt{164}$, $BC=\sqrt{656}=2AB$). $\alpha+2\beta=14+28=42$. Answer (B).