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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

  1. A \(2 x+2 y=1-\sqrt{6}\)
  2. B \(x+y=3-3 \sqrt{6}\)
  3. C \(x+y=2-\sqrt{6}\)
  4. D \(x+y=3-2 \sqrt{6}\)
Verified Solution

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…