Answer to Question #96884 in Analytic Geometry for Aakash

Question

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

Answer 1

The equation of a straight line with gradient m, passing through the point $(x_1, y_1)$, is 

.

$$y − y_1 = m(x − x_1).$$

Point  $A(1,√3)$ :

$$y − \sqrt{3}= m_1(x − 1 ).$$

The equation of the second line

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

Or

$$y = -\frac{1}{\sqrt{3}} (x+\sqrt{3}) \Rightarrow m_2=-\frac{1}{\sqrt{3}}$$

The angle between the lines:

$$\tan(\alpha)= |\frac{m_2-m_1}{1-m_2m_1} |$$

then

$\tan(\frac{ \pi }{6} )=|\frac{-\frac{1}{\sqrt{3}}-m_1}{1+\frac{1}{\sqrt{3}}m_1} | \Rightarrow \frac{1}{\sqrt{3}}=| \frac{-\frac{1}{\sqrt{3}}-m_1}{1+\frac{1}{\sqrt{3}}m_1} | \Rightarrow m_1=0 ,$ or $m_1=-\frac{\sqrt{3}}{2}$

The sought equations:

$$1) y=\sqrt{3} .$$

.

and

$$y − \sqrt{3}= -\frac{\sqrt{3}}{2} (x − 1 ) \Rightarrow$$

$$2) y = -\frac{\sqrt{3}}{2} x +\frac{3\sqrt{3}}{2}$$