Question

How do you write the equation given $\left( -6,7 \right)$; parallel to $3x+7y=3$?

Answer 1

Hint: In this problem we need to calculate the equation of the line which is parallel to the given line and passing through the given point. For this problem we are going to use the geometry rule that two parallel lines should have equal slope. So, we will consider the slope of the given line as ${{m}_{1}}$ and the slope of the required line as ${{m}_{2}}$. We know that the slope of the line which is in the form of $ax+by+c=0$ is $-\dfrac{a}{b}$. From this we can have the value of ${{m}_{1}}$ as well as ${{m}_{2}}$. Now we have the slope of the required line a point through which the line passes, so we will use the slope point formula and substitute all the values we have and simplify the equation to get the required solution.

Complete step-by-step solution:
Given line is $3x+7y=3$.
Comparing the above equation with stand equation which is $ax+by+c=0$, then we will get
$a=3$, $b=7$, $c=-3$.
Let us assume the slope of the given line as ${{m}_{1}}$ and slope of the required line as ${{m}_{2}}$.
Now the slope of the given line is calculated by
$\begin{align}
  & \Rightarrow {{m}_{1}}=-\dfrac{a}{b} \\
 & \Rightarrow m=-\dfrac{3}{7} \\
\end{align}$
In geometry we can say that the slopes of two parallel lines are equal to each other. So, the slope of the required line will be ${{m}_{2}}={{m}_{1}}=-\dfrac{3}{7}$. Now we have the slope of the required line and it passes through the point $\left( -6,7 \right)$. From slope point equation we are going to write the equation of the line as
$\begin{align}
  & \Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\
 & \Rightarrow y-7=-\dfrac{3}{7}\left( x-\left( -6 \right) \right) \\
 & \Rightarrow 7\left( y-7 \right)=-3\left( x+6 \right) \\
 & \Rightarrow 7y-49=-3x-18 \\
 & \Rightarrow 3x+7y=31 \\
\end{align}$
Hence the equation of the required line is $3x+7y=31$.

Note: We can follow another simple method to solve this problem. In geometry we can write the equation of the line which is parallel to the line $ax+by=c$ as $ax+by=d$. We will substitute the point we have in the equation $ax+by=d$ and calculate the $d$ value and substitute this value in the equation $ax+by=d$ to get the result.
We can write the equation of the line which is parallel to the line $3x+7y=3$ as $3x+7y=d$.
Substituting the point $\left( -6,7 \right)$ in the equation $3x+7y=d$, then we will get
$\begin{align}
  & \Rightarrow 3\left( -6 \right)+7\left( 7 \right)=d \\
 & \Rightarrow d=49-18 \\
 & \Rightarrow d=31 \\
\end{align}$
Substituting the $d$ value in $3x+7y=d$, then we will get the equation of the required line as $3x+7y=31$.
From both the methods we got the same result.