Question

The equation of the line passing through the point of intersection of the lines $4x-3y-1=0$ and $5x-2y-3=0$ and parallel to $2y-3x+2=0$ is
A. \[x-3y=1\]
B. \[3x-2y=1\]
C. \[2x-3y=1\]
D. \[2x-y=1\]

Answer 1

Hint: In this question, we are to find the equation of a line that passes through the point of intersection of the given lines and parallel to the line with equation $2y-3x+2=0$. We know from the coordinate geometry of straight lines, the slopes of two non-vertical parallel lines are equal. Thus, the slope of the given equation and the given point are applied in the point-slope form of the line. By substituting the given values, the required equation is obtained.



Formula Used:The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and parallel to the equation of the line $ax+by+c=0$ is $a(x-{{x}_{1}})+b(y-{{y}_{1}})=0$
The slope of a line with the equation in the form of $ax+by+c=0$ is $m=\dfrac{-a}{b}$
The equation of the line (point-slope form) is
$y-{{y}_{1}}=m(x-{{x}_{1}})$
For two non-vertical parallel lines, slopes are equal i.e., ${{m}_{1}}={{m}_{2}}$
The point of intersection of the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ is calculated by
\[\left( \dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\]



Complete step by step solution:It is given that,
A line that is passing through the point of intersection of the lines
$\begin{align}
  & 4x-3y-1=0\text{ }...(1) \\
 & 5x-2y-3=0\text{ }...(2) \\
\end{align}$
From (2),
$\begin{align}
  & 5x-2y-3=0 \\
 & \Rightarrow 2y=5x-3 \\
 & \Rightarrow y=\dfrac{5x}{2}-\dfrac{3}{2}\text{ }...(3) \\
\end{align}$
Then, substituting (3) in (1), we get
$\begin{align}
  & 4x-3y-1=0 \\
 & \Rightarrow 4x-3\left( \dfrac{5x}{2}-\dfrac{3}{2} \right)-1=0 \\
 & \Rightarrow 4x-\dfrac{15x}{2}+\dfrac{9}{2}-1=0 \\
 & \Rightarrow \dfrac{8x-15x}{2}=1-\dfrac{9}{2} \\
 & \Rightarrow -\dfrac{7x}{2}=-\dfrac{7}{2} \\
 & \text{ }\therefore x=1 \\
\end{align}$
Then, substituting in (3), we get
$\begin{align}
  & y=\dfrac{5x}{2}-\dfrac{3}{2} \\
 & \text{ }=\dfrac{5(1)}{2}-\dfrac{3}{2} \\
 & \text{ }=\dfrac{2}{2} \\
 & \therefore y=1 \\
\end{align}$
Therefore, the point of intersection of the given lines is $({{x}_{1}},{{y}_{1}})=(1,1)$.
The equation of the given parallel line is $2y-3x+2=0$.
Since we know that, the slopes of parallel lines are equal, the slope of the required line is
$m=\dfrac{-(-3)}{2}=\dfrac{3}{2}$
Then, the equation of the line passing through the point $(1,1)$ with a slope $m=\dfrac{3}{2}$ is
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & \Rightarrow y-1=\dfrac{3}{2}(x-1) \\
 & \Rightarrow 2(y-1)=3(x-1) \\
 & \Rightarrow 3(x-1)-2(y-1)=0 \\
\end{align}$
$\begin{align}
  & \Rightarrow 3x-3-2y+2=0 \\
 & \Rightarrow 3x-2y-1=0 \\
 & \therefore 3x-2y=1 \\
\end{align}$



Option ‘B’ is correct



Note: Here we may go wrong with the slope of the line. We need to remember that the slopes of two parallel lines are equal. And here we may also go wrong with the point of intersection of the two lines. We can also calculate the point of intersection by the direct formula we have. This type of question also can be solved by the standard equation of the parallel lines as mentioned above.