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
- A \(x + y = 2 -\sqrt 6\)
- B \(2x +2y = 1 -\sqrt 6\)
- C \(x + y = 3 -3\sqrt 6\)
- D \(x + y = 3 -2\sqrt 6\)
Answer & Solution
Correct Answer
(D) \(x + y = 3 -2\sqrt 6\)
Step-by-step Solution
Detailed explanation
\(x-y=4\) 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 \(2\sqrt 3 \) \(2\sqrt 3 = \frac{{\left| {2 + 1 - c} \right|}}{{\sqrt 2 }}\)…
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