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 $y=\sqrt{3}x$