Question: Find the equation of a straight line passing through (2,-3) and parallel to the line 14x +13y +9 = 0.

Solution:

Here, equation of line 1 is (14x +13y +9=0)
Slope of line 1 ($m_1$) = $\dfrac{- coefficient \; of \; x}{coefficient \; of \; y}$
$= \dfrac{-14}{13}$

We know, if two lines are parallel then their slopes are equal. Let $m_2$ be the slope of the other line then,
$m_1 = m_2$
$or, m_2 = m_1$
$\therefore m_2 = \frac{-14}{13}$

For line 2, we have coordinate of a single point (2,-3). Let (2,-3) be ($x_1,y_1$).
We know, slope of the line ($m_2$) is $\frac{-14}{13}$. Using formula, equation of line 2 is,

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

$or, y -(-3) =  \dfrac{-14}{13} × (x -2)$

$or, y +3 = \dfrac{-14(x-2)}{13}$

$or, 13(y+3) = -14x +28$

$or, 13y +39 = -14x +28$

$or, 14x +13y +39-28 = 0$

$or, 14x +13y +11= 0$ is the required equation.


Hence, the required equation of the line passing through (2,-3) and parallel to the line represented by (14x +13y +9=0) is (14x +13y +11 =0).