Question

Find the equations of two straight lines which are parallel to $x+7y+2=0$ and at $\sqrt{2}$ distance away from it.

Answer 1

Hint: We start solving this problem by first considering the given equation of the straight line $x+7y+2=0$ . Then we start finding the equations of two lines parallel to the given line $x+7y+2=0$ and at a distance of $\sqrt{2}$ units from it. We consider the formula, the equations of the lines parallel to the line $ax+by+c=0$ are of the form $ax+by+k=0$ where $k$ is some constant. Then we use the formula, the distance between two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is given by $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ . Then we equate the obtained distance to $\sqrt{2}$ to get the value of the constant term in the equations of required lines. Hence, we get the equations of two straight lines which are parallel to $x+7y+2=0$ and at $\sqrt{2}$ distance away from it.

Complete step by step answer:
Let us consider the equation of the given straight line, $x+7y+2=0$
Now, we find the equations of the lines parallel to the line $x+7y+2=0$.
Let us consider the formula, the equations of the lines parallel to the line $ax+by+c=0$ are of the form $ax+by+k=0$ where $k$ is some constant.
By using the above formula, the equations of the lines parallel to the line $x+7y+2=0$ are of the form $x+7y+k=0$ .
Now, we find the distance between the parallel lines $x+7y+2=0$ and $x+7y+k=0$.
Let us consider the formula, the distance between two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ is given by $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ .
By using the above formula, the distance between parallel lines $x+7y+2=0$ and $x+7y+k=0$ is given by
$\begin{align}
  & \dfrac{\left| 2-k \right|}{\sqrt{{{1}^{2}}+{{7}^{2}}}} \\
 & =\dfrac{\left| 2-k \right|}{\sqrt{1+49}} \\
 & =\dfrac{\left| 2-k \right|}{\sqrt{50}} \\
 & =\dfrac{\left| 2-k \right|}{\sqrt{25\times 2}} \\
 & =\dfrac{\left| 2-k \right|}{5\sqrt{2}} \\
\end{align}$
But, we were given that the distance between the parallel lines is $\sqrt{2}$ .
So, we get,
$\begin{align}
  & \dfrac{\left| 2-k \right|}{5\sqrt{2}}=\sqrt{2} \\
 & \Rightarrow \left| 2-k \right|=5\sqrt{2}\times \sqrt{2} \\
 & \Rightarrow \left| 2-k \right|=5\times 2 \\
 & \Rightarrow \left| 2-k \right|=10 \\
 & \Rightarrow 2-k=\pm 10 \\
\end{align}$
First, let us consider, $2-k=10$ , we get,
 $\begin{align}
  & 2-k=10 \\
 & \Rightarrow k=2-10 \\
 & \Rightarrow k=-8 \\
\end{align}$
Then we consider, $2-k=-10$ , we get,
$\begin{align}
  & 2-k=-10 \\
 & \Rightarrow k=2+10 \\
 & \Rightarrow k=12 \\
\end{align}$
So, we got,
$k=-8$ and $k=12$
Therefore, the equations of the two straight lines which are parallel to $x+7y+2=0$ and at $\sqrt{2}$ distance away from it are $x+7y-8=0$ and $x+7y+12=0$ .
Hence, the required equations of the straight lines are $x+7y-8=0$ and $x+7y+12=0$ and can be plotted as below.

Note:
The possibilities for making mistakes in this type of problems are, one may make a mistake by considering the formula of the distance between two parallel lines $ax+by+{{c}_{1}}=0$ and $ax+by+{{c}_{2}}=0$ as $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{{{a}^{2}}+{{b}^{2}}}$.