JEE Mains · Maths · STD 11 - 9. straight line
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 \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{E}\) is mid point of diagonals \(\begin{array}{ll}\frac{\alpha+\gamma}{2}=\frac{1+1}{2} & \& \frac{\beta+\delta}{2}=\frac{2+0}{2} \\ \alpha+\gamma=2 & \beta+\delta=2 \\ 2(\alpha+\beta+\gamma+\delta)=2(2+2)=8\end{array}\)
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