JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Straight Lines

$△ABC$ is isosceles with $AB=AC$ and $\angle A=\frac{2\pi }{3}=120^{\circ }$, so the base angles are $\angle B=\angle C=30^{\circ }$.

By the sine rule, $\frac{BC}{\sin A}=\frac{AB}{\sin C}$:


Since $B$ lies on the positive x-axis at distance $AB=4$ from $A(-1,0)$: $B=(3,0)$.

The base angle at $B$ is $30^{\circ }$, measured from $BA$ (which points in the negative x-direction from $B$); this places line $BC$ at slope $-\frac{1}{\sqrt{3}}$ through $B$:


Intersecting with $y=x+3$, i.e. $x=y-3$:


So $\beta =\frac{6}{1+\sqrt{3}}$ and $\alpha =\beta -3=\frac{6}{1+\sqrt{3}}-3=\frac{6-3(1+\sqrt{3})}{1+\sqrt{3}}=\frac{3(1-\sqrt{3})}{1+\sqrt{3}}=-\frac{6}{(1+\sqrt{3}{)}^2}$
(rationalising: $\frac{3(1-\sqrt{3})}{1+\sqrt{3}}\cdot \frac{1-\sqrt{3}}{1-\sqrt{3}}=\frac{3(1-\sqrt{3}{)}^2}{1-3}=\frac{3(4-2\sqrt{3})}{-2}=-6+3\sqrt{3}$, matching $-\frac{6}{(1+\sqrt{3}{)}^2}$ since $(1+\sqrt{3}{)}^2=4+2\sqrt{3}$).

This gives the clean relation $\alpha =-\frac{\beta }{1+\sqrt{3}}$, i.e. $\beta =-\alpha (1+\sqrt{3})$, so ${\beta }^2={\alpha }^2(1+\sqrt{3}{)}^2$ and ${\beta }^4={\alpha }^4(1+\sqrt{3}{)}^4$.



Since $\alpha =-\frac{6}{(1+\sqrt{3}{)}^2}$, ${\alpha }^2=\frac{36}{(1+\sqrt{3}{)}^4}$, so