Question

The angle between the lines $kx+y+9=0$, $y-3x=4$ is ${45}^o,$ then the value of $k$ is :

• A. $2$ only
• B. $2$ or $-\frac{1}{2}$
• C. $-2$ only
• D. $-2$ or $-\frac{1}{2}$

Answer 1

The correct option is B $2$ or $-\frac{1}{2}$

$kx+y+9=0$ and $\theta ={45}^{\circ }$

$slope=m_1=-k$

And $y-3x=4$

$slope=m_2=3$

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

$\tan \theta =\left|\frac{k+3}{1-3k}\right|$

$\tan {45}^{\circ }=1=\left|\frac{k+3}{1-3k}\right|$

$\Rightarrow 1-3k=k+3$ or $3k-1=k+3$

$\Rightarrow 4k=-2$ or $2k=4$

$k=\frac{-1}{2}$ or $k=2$