Answer to Question #258025 in Analytic Geometry for RiB

Question

Find the equation of the line which passes through the point (1,√3) and makes an angle of 30◦ with the line x-√3y+√3 = 0.

Answer 1

Write the equation of the line in slope-intercept form

$y=mx+b$

Line

$x-\sqrt{3}y+\sqrt{3}=0$ $y=\dfrac{1}{\sqrt{3}}x+1$ $slope_1=m_1=\dfrac{1}{\sqrt{3}}$

The angle between two lines

$\tan \theta=\pm\dfrac{m_2-m_1}{1+m_1m_2}$

Given $\theta=30\degree$

$\tan 30\degree=\pm\dfrac{m_2-m_1}{1+m_1m_2}$ $\dfrac{1}{\sqrt{3}}=\pm\dfrac{m_2-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}m_2}$ $\sqrt{3}+m_2=\pm(3m_2-\sqrt{3})$

Let

$\sqrt{3}+m_2=-(3m_2-\sqrt{3})$ $m_2=0$

The equation of the line which passes through the point $(1, \sqrt{3})$ is

$y=\sqrt{3}$

Let

$\sqrt{3}+m_2=(3m_2-\sqrt{3})$ $m_2=\sqrt{3}$

Point $(1, \sqrt{3})$

$\sqrt{3}=\sqrt{3}(1)+b=>b=0$

The equation of the line which passes through the point $(1, \sqrt{3})$ is

$y=\sqrt{3}x$

$y=\sqrt{3}$ or $y=\sqrt{3}x$