Question

How to determine the equation of the line parallel to $3x-2y+4=0$ and passing through $\left( 1,6 \right)$ ?

Answer 1

Firstly, here we find the slope for the given linear line equation. Let us then name it ${{m}_{1}}$ and the slope for the line equation which we need to find be ${{m}_{2}}$ . If they are parallel to each other, then we need to know that parallel lines check for the condition, ${{m}_{1}}={{m}_{2}}$ .Now that we know the slope, substitute the given coordinates inline equation to get the required linear equation.

Complete step-by-step solution:
The given line equation is, $3x-2y+4=0$
To check if a pair of lines are parallel or perpendicular or neither we must first find their slopes.
Any straight line can be written in slope-intercept form, $y=mx+c$
Where $m$ is said to be the slope of the line $\left( m=\tan \theta \right)$
And $c$ is the y-intercept.
Now writing the equation in slope-intercept form to find the slope.
$\Rightarrow 3x-2y+4=0$
$\Rightarrow 2y=3x+4$
Now isolate the $y$ variable.
$\Rightarrow y=\dfrac{3x+4}{2}$
Now separate the terms.
$\Rightarrow y=\dfrac{3x}{2}+2$
Comparing it with the slope-equation, $y=mx+b$
$\Rightarrow {{m}_{1}}=\dfrac{3}{2};b=2$
Now we know that parallel lines have the same slope.
Hence, ${{m}_{2}}=\dfrac{3}{2}$
We also know that the required line passes through $\left( 1,6 \right)$
The general form of a line equation is, $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$
Upon substituting the values, we get,
$\Rightarrow \left( y-6 \right)=\dfrac{3}{2}\left( x-1 \right)$
Now evaluate further.
$\Rightarrow \left( 2y-12 \right)=3\left( x-1 \right)$
$\Rightarrow 2y-12=3x-3$
Now group all the terms together.
$\Rightarrow 3x-2y+9=0$
Hence the required line equation is, $3x-2y+9=0$

Note: Whenever the slope of a line $m$ is $\infty$ it indicates that the equation is a straight line parallel to the $y$ axis. If the slope of the line $m$is $0$, then it indicates that the equation is a straight line parallel to the $x$ axis. Slope is also known as the “gradient”.