JEE Mains · Maths · STD 11 - 9. straight line
A line passing through the point \(A(9,0)\) makes an angle of \(30^{\circ}\) with the positive direction of \(\mathrm{x}\)-axis. If this line is rotated about \(A\) through an angle of \(15^{\circ}\) in the clockwise direction, then its equation in the new position is
- A \(\frac{y}{\sqrt{3}-2}+x=9\)
- B \(\frac{x}{\sqrt{3}-2}+y=9\)
- C \(\frac{x}{\sqrt{3}+2}+y=9\)
- D \(\frac{y}{\sqrt{3}+2}+x=9\)
Answer & Solution
Correct Answer
(A) \(\frac{y}{\sqrt{3}-2}+x=9\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Eq}^2: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)\)
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