A straight line $x - 2 y - 4 = 0$ is shifted parallel to it by 3 units away from the origin and then rotated by an angle of $30 °$ in the anticlockwise direction. If the slope of the new line formed m, then the integral part of ‘m’ is ……

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Detailed Solution

Slope of given line =1/2. Slope of required line = m

Now $\tan 30 ° = |\frac{m - \frac{1}{2}}{1 + m \frac{1}{2}}| \Rightarrow |\frac{2 m - 1}{2 + m}| = \frac{1}{\sqrt{3}} \Rightarrow \frac{2 m - 1}{2 + m} = \pm \frac{1}{\sqrt{3}}$

If $\frac{2 m - 1}{2 + m} = \frac{1}{\sqrt{3}} \Rightarrow 2 \sqrt{3} m - \sqrt{3} = 2 + m \Rightarrow (2 \sqrt{3} - 1) m = 2 + \sqrt{3}$

$\Rightarrow m = \frac{2 + \sqrt{3}}{2 \sqrt{3} - 1} = \frac{(2 + \sqrt{3}) (2 \sqrt{3} + 1)}{12 - 1} = \frac{8 + 5 \sqrt{3}}{11} = \frac{16.5}{11} = 1.5$