A straight line x-2y-4=0 is shifted parallel | vedclass

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(C) The original line is $x-2y-4=0$. The slope of this line is $m_1 = 1/2$.
When a line is shifted parallel to itself,its slope remains unchanged. Th

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The original line is $x-2y-4=0$. The slope of this line is $m_1 = 1/2$.
When a line is shifted parallel to itself,its slope remains unchanged. Therefore,the slope of the line after shifting remains $m_1 = 1/2 = 	an \alpha$,where $\alpha = 	an^{-1}(1/2)$.
After rotating the line by an angle of $30^{\circ}$ in the anti-clockwise direction,the new slope $m$ is given by the formula for the tangent of the sum of angles:
$m = 	an(\alpha + 30^{\circ}) = \frac{	an \alpha + 	an 30^{\circ}}{1 - 	an \alpha 	an 30^{\circ}}$
Substituting the values $	an \alpha = 1/2$ and $	an 30^{\circ} = 1/\sqrt{3}$:
$m = \frac{1/2 + 1/\sqrt{3}}{1 - (1/2)(1/\sqrt{3})} = \frac{(\sqrt{3}+2)/(2\sqrt{3})}{(\sqrt{3}-1)/(2\sqrt{3})} = \frac{\sqrt{3}+2}{\sqrt{3}-1}$
Rationalizing the denominator:
$m = \frac{(\sqrt{3}+2)(\sqrt{3}+1)}{3-1} = \frac{3+\sqrt{3}+2\sqrt{3}+2}{2} = \frac{5+3\sqrt{3}}{2}$
Using $\sqrt{3} \approx 1.732$:
$m \approx \frac{5+3(1.732)}{2} = \frac{5+5.196}{2} = \frac{10.196}{2} = 5.098$
The integral part of $m$ is $5$.

$A$ straight line $x-2y-4=0$ is shifted parallel to it by $3$ units away from the origin and then rotated by an angle of $30^{\circ}$ in the anti-clockwise direction. If the slope of the new line formed is $m$,then the integral part of $m$ is

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