Find the equation of the line passing through the intersection of the lines 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

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Solution

The point of intersection of lines 2x + y = 5 and x + 3y + 8 = 0 is given by
$(\frac{23}{5}, - \frac{21}{5})$
Now, the slope of the line 3x + 4y = 7 $or y = - \frac{3}{4} x + \frac{7}{4}$ is $- \frac{3}{4}$
Now, we know that the slopes of two parallel lines are equal.
So, the slope of the required line is $- \frac{3}{4}$
Now, the equation of the required line passing through $(\frac{23}{5}, - \frac{21}{5})$ and having slope $- \frac{3}{4}$ is given by
$y + \frac{21}{5} = - \frac{3}{4} (x - \frac{23}{5}) \Rightarrow y + \frac{21}{5} = - \frac{3}{4} x + \frac{69}{20} \Rightarrow y + \frac{3}{4} x = \frac{69}{20} - \frac{21}{5} \Rightarrow 20 y + 15 x = - 15 \Rightarrow 3 x + 4 y + 3 = 0$

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