AP EAMCET · Maths · Straight Lines
The point \(\mathrm{P}(2,1)\) is translated to a point \(\mathrm{Q}\) parallel to the line \(L \equiv x-y-4=0\) by \(2 \sqrt{3}\) units. If the point \(Q\) lies in the third quadrant, then the equation of the line passing through \(\mathrm{Q}\) and perpendicular to \(\mathrm{L}\) is
- A \(2 x+2 y=1-\sqrt{6}\)
- B \(x+y=3-3 \sqrt{6}\)
- C \(x+y=2-\sqrt{6}\)
- D \(x+y=3-2 \sqrt{6}\)
Answer & Solution
Correct Answer
(D) \(x+y=3-2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Given \(L \equiv x-y-4=0\) \[ \Rightarrow \frac{x}{(-4)}+\frac{y}{(4)}=1 \] Slope \(m_1\) of line ' \(L\) ' is \(m_1=1\) Since line ' \(L\) ' and line \(P Q\) is parallel to each other. Hence slope of line ' \(L\) ' \(=\) Slope of line \(P Q\) Hence equation of \(P Q\) can be…
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