Question: Find the equation of a straight line perpendicular to the straight line 6x-4y+14=0 and passing through (-3,3).

Solution:

Here, equation of line 1 is 6x-4y+14=0.
Slope of line 1 ($m_1$) = $\dfrac{- coefficient\; of\;x}{coefficient\; of \;y}$
$= \dfrac{-6}{-4}$
$= \dfrac{3}{2}$

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,
$m_1 × m_2 = -1$
$or, \dfrac{3}{2} × m_2 = -1$
$or, \dfrac{3 m_2}{2} = -1$
$or, 3 m_2 = -2$
$\therefore m_2 = \dfrac{-2}{3}$

Also, the other line (line 2) passes through the point (-3,3) = ($x_1,y_1$). Its slope is $m_2$ = \frac{-2}{3}$.

So, equation of line 2 is given by,

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

$or, y -3 = \dfrac{-2}{3} × \{x - (-3)\}$

$or, 3(y -3) = -2(x +3)$

$or, 3y - 9 = -2x -6$

$or, 2x +3y -9+6 = 0$

$or, 2x +3y -3 = 0$ is the required equation.

Therefore, the required equation of the straight line is 2x +3y -3 = 0.