Question

If the angle between the lines $kx-y+6=0,3x-5y+7=0$ is $\frac{\pi }{4},$ then one of the value of $k=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$kx-y+6=0$

$slope=m_1=\frac{-k}{-1}=k$

And $3x-5y+7=0$

$slope=m_2=\frac{-3}{-5}=\frac{3}{5}$

So, angle between this two lines is $\frac{\pi }{4}$

$\therefore tan\frac{\pi }{4}=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$1=\left|\frac{k-\frac{3}{5}}{1+\frac{3k}{5}}\right|$

$\therefore 1+\frac{3k}{5}=k-\frac{3}{5}$

$\frac{8}{5}=\frac{2k}{5}$

$k=4$