 # Find the equation of the line passing `
Question:

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

Solution:

Given lines are

2x + y = 5 ……1

x + 3y = -8 ……2

Firstly, we find the point of intersection of equation 1 and equation 2

Multiply the equation 2 by 2, we get

2x + 6y = -16 …….3

On subtracting equation 3 from 1, we get

2x + y – 2x – 6y = 5 – (-16)

On simplifying we get

⇒ -5y = 5 + 16

⇒ -5y = 21

$\Rightarrow y=-\frac{21}{5}$

Putting the value of $y$ in equation 1 , we get

$2 x+\left(-\frac{21}{5}\right)=5$

On rearranging we get

$\Rightarrow 2 x=5+\frac{21}{5}$

$\Rightarrow 2 \mathrm{x}=\frac{25+21}{5}$

$\Rightarrow 10 x=46$

$\Rightarrow \mathrm{x}=\frac{46}{10}=\frac{23}{5}$

Hence, the point of intersection is $\left(\frac{23}{5},-\frac{21}{5}\right)$

Now, we find the slope of the given equation $3 x+4 y=7$ We know that the slope of an equation is

$m=-a / b$

$\Rightarrow m=-\frac{3}{4}$

So, the slope of a line which is parallel to this line is also $-\frac{3}{4}$

Then the equation of the line passing through the point $\left(\frac{23}{5},-\frac{21}{5}\right)$ having

slope $^{-\frac{3}{4}}$ is:

$y-y_{1}=m\left(x-x_{1}\right)$

Substituting the values we get

$\Rightarrow y-\left(-\frac{21}{5}\right)=-\frac{3}{4}\left(x-\frac{23}{5}\right)$

Computing and simplifying

$\Rightarrow y+\frac{21}{5}=-\frac{3}{4} x+\frac{69}{20}$

$\Rightarrow \frac{3}{4} \mathrm{x}+\mathrm{y}=\frac{69}{20}-\frac{21}{5}$

$\Rightarrow \frac{3 x+4 y}{4}=\frac{69-84}{20}$

$\Rightarrow 3 x+4 y=-\frac{15}{5}$

$\Rightarrow 3 x+4 y+3=0$

$\Rightarrow \frac{3 x+4 y}{4}=\frac{69-84}{20}$

$\Rightarrow 3 x+4 y=-\frac{15}{5}$

$\Rightarrow 3 x+4 y+3=0$