WBJEE · Maths · Straight Lines
A line through the point \(\mathrm{A}(2,0)\) which makes an angle of \(30^{\circ}\) with the positive direction of \(x\)-axis is rotated about A in clockwise direction through an angle \(15^{\circ}\). Then the equation of the straight line in the new position is
- A \((2-\sqrt{3}) x+y-4+2 \sqrt{3}=0\)
- B \((2-\sqrt{3}) x-y-4+2 \sqrt{3}=0\)
- C \((2-\sqrt{3}) x-y+4+2 \sqrt{3}=0\)
- D \((2-\sqrt{3})_x+y+4+2 \sqrt{3}=0\)
Answer & Solution
Correct Answer
(B) \((2-\sqrt{3}) x-y-4+2 \sqrt{3}=0\)
Step-by-step Solution
Detailed explanation
Hints : Equation of line in new position : \[ \begin{aligned} & y-0=\tan 15^{\circ}(x-2) \\ & \Rightarrow y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-2) \\ & \Rightarrow y=\frac{(\sqrt{3}-1)^2}{2}(x-2) \end{aligned} \]…
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