MHT CET · Maths · Straight Lines
If the line joining two points \(\mathrm{A}(2,0)\) and \(\mathrm{B}(3,1)\) is rotated about \(\mathrm{A}\) in anticlockwise direction through an angle of \(15^{\circ}\), then the equation of the line in new position is
- A \(y=3 x-6\)
- B \(y=\sqrt{3} x-2 \sqrt{3}\)
- C \(y=-\sqrt{3} x+2 \sqrt{3}\)
- D \(y=\frac{1}{\sqrt{3}} x-\frac{2}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(y=\sqrt{3} x-2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Refer Figure
Slope of \(\mathrm{AB}=\frac{1-0}{3-2}=1\)
\(\therefore \tan \theta=1 \Rightarrow \theta=45^{\circ}\)
Line \(\mathrm{AB}\) is rotated through \(15^{\circ}\) in anticlockwise direction about
A.
Therefore in a new position, slope of line \(=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}\) and it passes through \(\mathrm{A}\).
Required equation of line is \((\mathrm{y}-0)=\sqrt{3}(\mathrm{x}-2) \Rightarrow\)
\(\sqrt{3} x-2 \sqrt{3}=y\)
Slope of \(\mathrm{AB}=\frac{1-0}{3-2}=1\)
\(\therefore \tan \theta=1 \Rightarrow \theta=45^{\circ}\)
Line \(\mathrm{AB}\) is rotated through \(15^{\circ}\) in anticlockwise direction about
A.
Therefore in a new position, slope of line \(=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}\) and it passes through \(\mathrm{A}\).
Required equation of line is \((\mathrm{y}-0)=\sqrt{3}(\mathrm{x}-2) \Rightarrow\)
\(\sqrt{3} x-2 \sqrt{3}=y\)
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