Question

Find the equation of line passing through the intersection of lines $x - 2y = 1$ and $x + 3y = 2$ and parallel to the line $3x + 4y = 0$.

Answer 1

Hint: First find the slope of the line using the equation of parallel line given in the question as we know that the slopes of parallel lines are equal. Then find the point of intersection of the other two lines by solving them simultaneously. Then satisfy this point with the general equation of line $y = mx + c$ to get the answer.

Complete step-by-step answer:
Let the equation of line is $y = mx + c$ and it is denoted as $L$.
$ \Rightarrow L:y = mx + c$
According to the question, the line is parallel to the other line $3x + 4y = 0$. So their slopes will be equal. Finding the slope of this given line, we have:
$
   \Rightarrow 4y = - 3x \\
   \Rightarrow y = - \dfrac{3}{4}x
 $
Thus the slope of the line $3x + 4y = 0$ is $ - \dfrac{3}{4}$ and hence the slope of line $L$ will also be $ - \dfrac{3}{4}$. So we have:
$ \Rightarrow m = - \dfrac{3}{4}$
Putting this value in the equation of line $y = mx + c$, we’ll get:
$ \Rightarrow y = - \dfrac{3}{4}x + c{\text{ }}.....{\text{(1)}}$
Further, it is given that the line is also passing through the point of intersection of lines $x - 2y = 1$ and $x + 3y = 2$.
For finding this point of intersection, we will solve these two equations:
$
  x - 2y = 1{\text{ }}.....{\text{(a)}} \\
  x + 3y = 2{\text{ }}.....{\text{(b)}}
 $
Subtracting equation (a) from equation (b), we’ll get:
$
   \Rightarrow x + 3y - x + 2y = 2 - 1 \\
   \Rightarrow 5y = 1 \\
   \Rightarrow y = \dfrac{1}{5}
 $
Putting this value in equation (a), we’ll get:
$
   \Rightarrow x - \dfrac{2}{5} = 1 \\
   \Rightarrow x = 1 + \dfrac{2}{5} = \dfrac{7}{5}
 $
So the point of intersection is $\left( {\dfrac{7}{5},\dfrac{1}{5}} \right)$. This point is also lying on $L$ as the line is passing through the point of intersection. Hence, on satisfying this point in equation (1), we’ll get:
$
   \Rightarrow \dfrac{1}{5} = - \dfrac{3}{4} \times \dfrac{7}{5} + c \\
   \Rightarrow c = \dfrac{1}{5} + \dfrac{{21}}{{20}} \\
   \Rightarrow c = \dfrac{{4 + 21}}{{20}} = \dfrac{{25}}{{20}} \\
   \Rightarrow c = \dfrac{5}{4}
 $
Putting this value back in equation (1), we’ll get:
$
   \Rightarrow y = - \dfrac{3}{4}x + \dfrac{5}{4} \\
   \Rightarrow 4y = - 3x + 5 \\
   \Rightarrow 3x + 4y = 5
 $

Thus the equation of the required line is $3x + 4y = 5$.

Note: If the slope of line and its y-intercept is given then the general equation of line that can be used is:
$ \Rightarrow y = mx + c$, where $m$ and $c$ are slope and y-intercept respectively.
If two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ lying on the line are known then the general equation that can be used is:
$ \Rightarrow \left( {y - {y_1}} \right) = \dfrac{{\left( {{y_2} - {y_1}} \right)}}{{\left( {{x_2} - {x_1}} \right)}}\left( {x - {x_1}} \right)$
And if both x and y intercepts are given and they are $a$ and $b$ respectively, then the general equation that can be used is:
$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} = 1$