Question

Consider two lines. Line 1 is given by the equation y=x and Line 2 is
represented by the equation v3y=x+√3.What is the angle between the two
lines?​

Answer 1

Answer:

In line one the equation is y=x

=then,y in right =_y and x=_x

In second y=x+/3_9xy

Answer 2

Given:

Line 1 is given by the equation y=x and Line 2 is

represented by the equation 3y=x+√3.

To find:

Angle between the lines.

Calculation:

Line 1 :

$y = x$

So, slope be m1 ;

$m1 = 1 \: \: \: \: \: \: ......(1)$

Line 2:

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

$= > y = \dfrac{x}{3} + \dfrac{ \sqrt{3} }{3}$

$= > y = \dfrac{x}{3} + \dfrac{ 1 }{ \sqrt{3} }$

So, let slope be m2;

$m2 = \dfrac{1}{3} \: \: \: \: \: .....(2)$

Let angle between lines be $\theta$.

$\tan( \theta) = | \dfrac{m1 - m2}{1 + (m1)(m2)} |$

$= > \tan( \theta) = | \dfrac{1 - \frac{1}{3} }{1 + (1)( \frac{1}{3} )} |$

$= > \tan( \theta) = | \dfrac{ \frac{2}{3} }{1 + \frac{1}{3} } |$

$= > \tan( \theta) = | \dfrac{ \frac{2}{3} }{ \frac{4}{3} } |$

$= > \tan( \theta) = \dfrac{1}{2}$

$= > \theta = { \tan}^{ - 1} ( \dfrac{1}{2} )$

So, final answer is:

$\boxed{ \bf{ \theta = { \tan}^{ - 1} ( \dfrac{1}{2} )}}$