Question: Find the equation of the straight lines which pass through the point (0,0) and making an angle of 60° with the line x+y+3=0.

Solution:

Line 1,
Equation of line is x +y +3 = 0
Slope ($m_1$) = $\frac{- coefficient \; of \; x}{coefficient \; of \; y}$
$= \frac{-1}{1}$
$= -1$

Given,
Angle between two straight lines ($\theta$) = 60°. Let the slope of the other line be represented by $m_2$.

Using formula,

$tan \theta = \left ( \pm \frac{m_1-m_2}{1 + m_1×m_2} \right )$

$or, tan 60° = \left ( \pm \frac{-1 - m_2}{1 + (-1)m_2} \right )$

$or, √3 = \left ( \pm \frac{-1 - m_2}{1 - m_2} \right )$

$or, √3 ( 1 - m_2) = \pm (-1 - m_2)$

Taking positive sign,
$or, √3(1 - m_2) = -1 - m_2$
$or, √3 - √3m_2 = -1 - m_2$
$or, √3 + 1 = √3 m_2 -m_2$
$or, (√3+1) = (√3-1)m_2$
$or, \frac{√3 +1}{√3 -1} = m_2$
$or, m_2 =  \frac{√3 +1}{√3 -1}$
$or, m_2 = \frac{√3 +1}{√3 -1} × \frac{√3 +1}{√3 +1}$
$or, m_2 = \frac{(√3+1)^2}{(√3)^2 - (√1)^2}$
$or, m_2 = \frac{(√3)^2 +2×√3×1 + 1^2}{3-1}$
$or, m_2 = \frac{3 + 2√3 +1}{2}$
$or, m_2 = \frac{4+2√3}{2}$
$or, m_2 = \frac{2(2+√3)}{2}$
$\therefore m_2 = 2+√3$

Taking negative sign,
$or, √3(1 - m_2) = -(-1-m_2)$
$or, √3 - √3 m_2 = 1 + m_2$
$or, √3 -1 = √3m_2 + m_2$
$or, (√3 -1) = (√3+1)m_2$
$or, \frac{√3-1}{√3+1} = m_2$
$or, m_2 = \frac{√3 -1}{√3+1} × \frac{√3-1}{√3-1}$
$or, m_2 = \frac{(√3-1)^2}{(√3)^2 - (√1)^2}$
$or, m_2 = \frac{(√3)^2 -2×√3×1 + 1^2}{3-1}$
$or, m_2 = \frac{3 -2√3 +1}{2}$
$or, m_2 = \frac{4-2√3}{2}$
$or, m_2 = \frac{2(2-√3)}{2}$
$\therefore m_2 = 2√3$

Now,

Equation of line 2 when $m_2 = 2+√3$ and point (0,0) = ($x_1,y_1$), 
Using formula,
$y - y_1 = m_2 (x -x_1)$
$or, y -0 = (2+√3) (x -0)$
$or, y = (2+√3)x$
$or, (2+√3)x - y = 0$ is the required equation.

Equation of line 2 when $m_2 = 2-√3$ and point (0,0) = ($x_1,y_1$), 
Using formula,
$y - y_1 = m_2 (x -x_1)$
$or, y -0 = (2-√3) (x -0)$
$or, y = (2-√3)x$
$or, (2-√3)x -y = 0$ is the required equation.

Therefore, the required equation of the straight lines are {(2+√3)x -y = 0} and {(2-√3)x - y = 0}.