Question: Find the equation of a straight line perpendicular to the line 5x+6y+4=0 and passing through (-9,0).

Solution:

Here, equation of line 1 is 5x+6y+4=0.
Slope of line 1 ($m_1$) = $\dfrac{- coefficient\; of\;x}{coefficient\; of \;y}$
$= \dfrac{-5}{6}$

Given, two lines i.e. line 1 and line 2 with slopes $m_1$ and $m_2$ respectively are perpendicular to each other.
We know, when two lines are perpendicular,
$m_1 × m_2 = -1$
$or, \dfrac{-5}{6} × m_2 = -1$
$or, \dfrac{-5 m_2}{6} = -1$
$or, -5 m_2 = -6$
$or, 5m_2 = 6$
$\therefore m_2 = \dfrac{6}{5}$

Also, the other line (line 2) passes through the point (-9,0) = ($x_1,y_1$). Its slope is $m_2$ = \frac{6}{5}$.

So, equation of line 2 is given by,

$y - y_1 = m_2 (x -x_1)$

$or, y -0 = \dfrac{6}{5} × \{x - (-9)\}$

$or, 5 y = 6(x +9)$

$or, 5y = 6x +54$

$or, 6x -5y +54 = 0$ is the required equation.

Therefore, the required equation of the straight line is 6x -5y +54 = 0.