Straight Lines Question 12
Question 12 - 2024 (31 Jan Shift 1)
Let $\alpha, \beta, \gamma, \delta \in \mathrm{Z}$ and let $\mathrm{A}(\alpha, \beta), \mathrm{B}(1,0), \mathrm{C}(\gamma, \delta)$ 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
$\mathrm{D}(1,2)$
$\mathrm{C}(\gamma, \delta)$
$\mathrm{A}(\alpha, \beta)$
$\mathrm{B}(1,0)$
Let $\mathrm{E}$ is mid point of diagonals
$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$
$& \frac{\beta+\delta}{2}=\frac{2+0}{2}$
$\alpha+\gamma=2$
$\beta+\delta=2$
$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$
