Q: Find the equation of the straight lines passing through (-2 , -7) and having intercept of length 3 units between the straight lines 4x + 3y = 12 and 4x + 3y = 3.

Sol: Distance between the two given parallel lines

$\displaystyle =\frac{c_1 – c_2}{\sqrt{a^2 + b^2}} = \frac{12 – 3}{\sqrt{16 + 9}} = \frac{9}{5} $

$\displaystyle tan\theta = \frac{9}{12} = \frac{3}{4}$

Slope of the parallel lines $\displaystyle =- \frac{4}{3} = m_2 $

Also , $\displaystyle tan\theta = \pm \frac{m_1 – m_2}{1 + m_1 m_2}$

$\displaystyle \frac{3}{4} = \pm \frac{m_1 + \frac{4}{3}}{1 – \frac{4}{3} m_1}$

$\displaystyle m_1 + \frac{4}{3} = \frac{3}{4} (1 – \frac{4}{3} m_1)$

and , $\displaystyle m_1 + \frac{4}{3} = -\frac{3}{4} (1 – \frac{4}{3} m_1)$

The slopes are m_{1} = -7/24 , m_{1} = ∞ (the line is parallel to the y – axis)

The required equations of the lines are 7x + 24y + 182 = 0 and x+2 =0.