JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Straight Lines

$A(4,6)$; bisector of angle $B$ is $y=x$; side $AC$: $2x-y=2$.

$D$ = intersection of the bisector with $AC$: substituting $y=x$ into $2x-y=2$ gives $D=(2,2)$.

By the angle-bisector theorem, $AD:DC=AB:BC$. Given $2AB=BC$, this ratio is $1:2$, so


Reflecting $A(4,6)$ about the bisector $y=x$ gives $A^{\prime }=(6,4)$; since reflecting one arm of the angle across its bisector maps it onto the other arm, $A^{\prime }$ lies on line $BC$. So line $BC$ passes through $C(-2,-6)$ and $A^{\prime }(6,4)$:


Intersecting with the bisector $y=x$:


So $B=(\alpha ,\beta )=(14,14)$.



Key confirmed: A.