ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 9. straight line

The point \((2, 1 )\)  is translated parallel to the line \(L\) \(: x - y= 4\) by \(2\sqrt 3\,units\) . If the new points \(Q\)  lies in the third quadrant, then the equation of the line passing through \(Q\) and perpendicular to \(L\)  is

  1. A \(x + y = 2 -\sqrt 6\)
  2. B \(2x +2y = 1 -\sqrt 6\)
  3. C \(x + y = 3 -3\sqrt 6\)
  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

\(x-y=4\) To find equation of \(R\) slop of \(L=0\) is \(1\) \( \Rightarrow \)slop of \(QR=-1\) Let \(QR\) is \(y=mx+c\) \(y=-x+c\) \(x+y-c=o\) distance of \(QR\) from \((2,1)\) is \(2\sqrt 3 \) \(2\sqrt 3 = \frac{{\left| {2 + 1 - c} \right|}}{{\sqrt 2 }}\)…