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AP EAMCET · Maths · Straight Lines

The diagonals \(\mathrm{AC}\) and \(\mathrm{BD}\) of a rhombus \(\mathrm{ABCD}\) intersect at the point \((3,4)\). If \(\mathrm{BD}=2 \sqrt{2}, \mathrm{~A}=(1,2), \mathrm{B}=(\alpha, \beta)\),\(D=(\gamma, \delta)\) and \(\alpha < \delta < \gamma < \beta\), then \(\beta+\gamma-\delta=\)

  1. A \(0\)
  2. B \(\alpha + 4\)
  3. C \(-2\alpha +6\)
  4. D \(-3\alpha + 12\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-3\alpha + 12\)

Step-by-step Solution

Detailed explanation

\(\because B D=2 \sqrt{2}\) \(\therefore O B=O D=\sqrt{2}\) \(\because \quad(3,4)\) is mid point of \(A C\) \(\therefore C=(5,6)\) So, \(O A=O C=\sqrt{4+4}=2 \sqrt{2}\) In \(\triangle A O B, O A^2+O B^2=A B^2\) \(\Rightarrow A B^2=10 \Rightarrow A B=\sqrt{10}\)…