Question

The line passing through $(7,3)$, $(5,1)$ meets the $x$-axis at $P$. lf the line is rotated through an angle ${30}^o$ in the anti-clockwise direction about $P$, then the slope of its new position is :

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $2+\sqrt{3}$
• D. $2-\sqrt{3}$

Answer 1

The correct option is D $2-\sqrt{3}$

Slop of line $=\frac{3-1}{7-5}=\frac{2}{2}=1\Rightarrow \theta ={45}^0$

$y-1=x-5\Rightarrow y=x-4$

It meets $x$-axis at $(4,0)$

$x=\alpha +x^1\cos \theta -y^1\sin \theta$

$y=\beta +x^1\sin \theta +y^1\cos \theta$

$\Rightarrow x=4+x^1\left(\frac{\sqrt{3}}{2}\right)-y^1\left(\frac{1}{2}\right)$

$y=0+x^1\left(\frac{1}{2}\right)+y^1\left(\frac{\sqrt{3}}{2}\right)$

$y=x-4$

$\Rightarrow \frac{x^1}{2}+\frac{y^1\sqrt{3}}{2}=4+\frac{\sqrt{3}x}{2}-\frac{y^1}{2}-4$

$\Rightarrow m=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}$