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  • If L1 has slope 1 and L2 is parallel to the line <img alt="x-\sqrt{3}y+1= 0" src="https://entrancecorner.oncodecogs.com/gif.latex?x-%5Csqrt%7B3%7Dy&plus;1%3

If L1 has slope 1 and L2 is parallel to the line x-\sqrt{3}y+1= 0, then the angle between L1 and L2 is

Option: 1

\tan^{-1}\left ( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right )


Option: 2

\frac{\pi}{4}


Option: 3

\tan^{-1}\left ( \frac{\sqrt{3}+1}{2} \right )


Option: 4

\tan^{-1}\left ( \sqrt{2}-1 \right )


Answers (1)

best_answer

Given slope of L_{1}= m_{1}= 1 

Now L2 is parallel to x-\sqrt{3}y+1= 0, so slope of L2(m2)  = = slope\: of\: x-\sqrt{3}y+1= 0

= \frac{1}{\sqrt{3}}

Angle between L_{1} and L_{2}
\tan \Theta= \left |\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}\cdot 1} \right |= \left | \frac{\sqrt{3}-1}{\sqrt{3}+1}\right |

Posted by

Shailly goel

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