JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Straight Lines

Since P(1,1) is the circumcentre, it is equidistant from all three vertices. The perpendicular from P to side AC is horizontal (since AC is vertical with both points having x-coordinate a), so the midpoint D of AC has coordinates $(a,\frac{b+3}{2})$. Since the perpendicular has slope 0, we get $\frac{b+3}{2}=1$, giving $b=-1$.

For side BC, the midpoint is $E\left(\frac{a+b}{2},\frac{5+b}{2}\right)=\left(\frac{a-1}{2},2\right)$. The perpendicularity condition $m_{BC}\cdot m_{PE}=-1$ gives $\frac{5-b}{b-a}\cdot \frac{2-1}{\frac{a-1}{2}-1}=-1$. Substituting $b=-1$ and simplifying: $\frac{6}{-1-a}\cdot \frac{2}{a-3}=-1$, leading to $a^2-2a-15=0$. This gives $a=5$ or $a=-3$. Since $ab>0$ and $b=-1$, we need $a<0$, so $a=-3$.

Line AP through A(-3,3) and P(1,1) has equation $x+2y=3$. Line BC through B(-1,5) and C(-3,-1) has equation $y=3x+8$. Solving these simultaneously: $x+2(3x+8)=3$ gives $7x=-13$, so $x=-\frac{13}{7}$ and $y=\frac{17}{7}$. Therefore $k_1+k_2=\frac{4}{7}$.