Question

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 intcgral part of ' $m$ ' is

• A. $-1$
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (C)

$x-2y-4=0$

(i) Then rotates $30^{\circ }$ angle in only clockwise direction.
$\therefore$ slope of new line is $m$
(ii) Line is shifted 3 units away from origin. So, equation will be $x-2y-7=0$
Now, it rotates $30^{\circ }$ angle in anticlockwise direction.
So,

	x	y
x'	$cos\theta$	$sin\theta$
y'	$-sin\theta$	$cos\theta$

So, $x=x^{\prime }\cos \theta -y^{\prime }\sin \theta$
So, $y=x^{\prime }\sin \theta +y^{\prime }\cos \theta$
$x=\frac{x^{\prime }\sqrt{3}}{2}-\frac{y^{\prime }}{2}$ and $y=\frac{x^{\prime }}{2}+\frac{\sqrt{3}{y}^{\prime }}{2}$
Now, equation will be
$\left[\frac{\sqrt{3}}{2}{x}^{\prime }-\frac{y^{\prime }}{2}\right]-2\left[\frac{x^{\prime }}{2}+\frac{\sqrt{3}{y}^{\prime }}{2}\right]-7=0$
$\Rightarrow \left(\frac{\sqrt{3}-2}{2}\right)x^{\prime }-\left(\frac{1+2\sqrt{3}}{2}\right)y^{\prime }=7$
So, $m=\frac{\sqrt{3}-2}{1+2\sqrt{3}}$
$\Rightarrow \frac{1.732-2}{1+2(1.732}$
So, integral part is (1)