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).
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