Question

if the lines $y=(2+\sqrt{3)}x+4andy=kx+6$ are inclined at an angle ${60}^{\circ }$ to each other, then 11 th value of k will be

• A. 1
• B. 2
• C. -1
• D. -2

Answer 1

Answer: (C)

-1

Given lines are $l_1:y=(2+\sqrt{3})x+4$ and $l_2:y=kx+6$

Slope of line $l_1:m_1=2+\sqrt{3}$ and slope of line $l_2:m_2=k$

The angle between two line is given by,

$tan\theta =|\frac{m_1-m_2}{1+m_1{m}_2}|$

$\Rightarrow tan60=|\frac{2+\sqrt{3}-k}{1+(2+\sqrt{3}k)}|$

$\Rightarrow \sqrt{3}=|\frac{2+\sqrt{3}-k}{1+2k+\sqrt{3}.k}|$

$\Rightarrow \sqrt{3}+3k+2\sqrt{3}k=\pm (2+\sqrt{3}-k)$

$(2\sqrt{3}+3)k+k=2$

$\Rightarrow k=\frac{2}{(2\sqrt{3}+4)}$

Now when $LHS=-\sqrt{3}$

$\frac{2+\sqrt{3}-k}{1+2k+\sqrt{3}.k}=-\sqrt{3}$

$\Rightarrow 2+\sqrt{3}-k=-\sqrt{3}-3k-2\sqrt{3}$

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

$\Rightarrow k=-1$