9.Straight Line
hard

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 $\mathrm{ABCD}$. If $\mathrm{AB}=\sqrt{10}$ and the points $\mathrm{A}$ and $\mathrm{C}$ lie on the line $3 y=2 x+1$, then $2(\alpha+\beta+\gamma+\delta)$ is equal to

A

$10$

B

$5$

C

$12$

D

$8$

(JEE MAIN-2024)

Solution

Let $\mathrm{E}$ is mid point of diagonals

$\begin{array}{ll}\frac{\alpha+\gamma}{2}=\frac{1+1}{2} & \& \frac{\beta+\delta}{2}=\frac{2+0}{2} \\ \alpha+\gamma=2 & \beta+\delta=2 \\ 2(\alpha+\beta+\gamma+\delta)=2(2+2)=8\end{array}$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.