Straight Lines Question 12
Question 12 - 2024 (31 Jan Shift 1)
Let $\alpha, \beta, \gamma, \delta \in Z$ and let $A(\alpha, \beta), B(1,0), C(\gamma, \delta)$ and $D(1,2)$ be the vertices of a parallelogram $A B C D$. If $A B=\sqrt{10}$ and the points $A$ and $C$ lie on the line $3 y=2 x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to
(1) 10
(2) 5
(3) 12
(4) 8
Show Answer
Answer (4)
Solution
$D(1,2)$
$C(\gamma, \delta)$
$A(\alpha, \beta)$
$B(1,0)$
Let $E$ is mid point of diagonals
$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$
$\& \frac{\beta+\delta}{2}=\frac{2+0}{2}$
$\alpha+\gamma=2$
$\beta+\delta=2$
$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$
