Question

The angle between the lines $\sqrt{3x}+y+1=0$ and $x+1=0$ is.

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{3}$

We have,

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

$x+1=y.......\left(2\right)$

From equation (1) to,

$\sqrt{3}x+y+1=0$

$y=-\sqrt{3}x-1$

On compare that,

$y=m_{1}x+C$

Now,

$m_1=-\sqrt{3}$

From equation (2) to

$x+1=0$

Now,

$m_2=-1$

Then,

Angle

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

$\tan \theta =\sqrt{3}$

$⟹\theta =\frac{\pi }{3}$