JEE Mains · Maths · STD 11 - 9. straight line
Let \(\mathrm{A}(-2,-1), \mathrm{B}(1,0), \mathrm{C}(\alpha, \beta)\) and \(\mathrm{D}(\gamma, \delta)\) be the vertices of a parallelogram \(A B C D\). If the point \(C\) lies on \(2 x-y=5\) and the point \(D\) lies on \(3 x-2 y=6\), then the value of \(|\alpha+\beta+\gamma+\delta|\) is equal to ...........
- A \(30\)
- B \(31\)
- C \(32\)
- D \(33\)
Answer & Solution
Correct Answer
(C) \(32\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right) \) \( \frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} \)…
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