JEE Main 2024 (Online) 31st January Morning ShiftQuestion 1
Let $\alpha,\beta,\gamma,\delta\in\mathbb{Z}$ and let $A(\alpha,\beta)$, $B(1,0)$, $C(\gamma,\delta)$ and $D(1,2)$ be the vertices of a parallelogram $ABCD$. If $AB=\sqrt{10}$ and the points $A$ and $C$ lie on the line $3y=2x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to
Points and Straight Lines
Solution
$A$ on $3y=2x+1$: $\beta=\frac{2\alpha+1}{3}$ with integer $\alpha$ ($\alpha\equiv1 \pmod 3$). $AB^2=(\alpha-1)^2+\beta^2=10$ gives $(\alpha,\beta)=(-2,-1)$ (others fail integrality or distance; e.g. $(4,3)$ gives $AB^2=18$). Diagonals bisect: $C=B+D-A=(2-\alpha,2-\beta)=(4,3)$, which lies on $3y=2x+1$ ($9=9$). $2(\alpha+\beta+\gamma+\delta)=2(-2-1+4+3)=8$. Answer (B).