JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Straight Lines

$A(3,-1)$, $B(-2,3)$, orthocentre $H(1,1)$.

Finding $C=(\alpha ,\beta )$: since $AH\perp BC$, slope of $AH=\frac{1-(-1)}{1-3}=-1$, so line $BC$ has slope $1$: $y-3=1(x+2)\Rightarrow y=x+5$.

Since $BH\perp AC$, slope of $BH=\frac{1-3}{1-(-2)}=-\frac{2}{3}$, so line $AC$ has slope $\frac{3}{2}$: $y+1=\frac{3}{2}(x-3)\Rightarrow y=\frac{3}{2}x-\frac{11}{2}$.

$C$ is the intersection of lines $BC$ and $AC$:

So $C=(21,26)$, giving $\alpha +\beta =47$.

Finding $(h,k)$, circumcentre of $△PAB$ ($P=H=(1,1)$): perpendicular bisector of $PA$ — midpoint $(2,0)$, slope of $PA=-1$, so bisector slope $1$: $y=x-2$.

Perpendicular bisector of $PB$ — midpoint $(-0.5,2)$, slope of $PB=\frac{3-1}{-2-1}=-\frac{2}{3}$, so bisector slope $\frac{3}{2}$: $y-2=\frac{3}{2}(x+0.5)\Rightarrow y=\frac{3}{2}x+\frac{11}{4}$.

Solving: $x-2=\frac{3}{2}x+\frac{11}{4}⟹-\frac{1}{2}x=\frac{19}{4}⟹x=-\frac{19}{2},y=-\frac{23}{2}$.

So $(h,k)=\left(-\frac{19}{2},-\frac{23}{2}\right)$, giving $2(h+k)=-42$.