Question

How do you write an equation of the line given $\left( 1,5 \right)$ that are parallel to and perpendicular to the line equation $y+4x=7$?

Answer 1

Change of form of the given equation $y+4x=7$ will give the slope of the line. We change it to the form of $y=mx+k$ to find the slope $m$. Then, we get into the condition which satisfies for a parallel and perpendicular line to the given equation $y+4x=7$. We apply those conditions and put the point value of $\left( 1,5 \right)$ to find the equations.

Complete answer:
The given equation $y+4x=7$ is of the form $ax+by=c$. Here a, b, c are the constants.
We convert the form to $y=mx+k$. $m$ is the slope of the line.
So, converting the equation we get
$\begin{aligned} & y+4x=7 \\\ & \Rightarrow y=-4x+7 \\\ \end{aligned}$
This gives that the slope of the line $y+4x=7$ is $-4$.
Now we have to find the equations of the lines which are going through $\left( 1,5 \right)$ being parallel and perpendicular to the given equation $y+4x=7$.
We know that the slope of a line perpendicular to a given line with slope $m$ will be always $-\dfrac{1}{m}$ and the slope of a line parallel to a given line with slope $m$ will be always $m$.
Therefore, the respective slopes for the parallel and perpendicular line to the given equation $y+4x=7$ will be $\dfrac{1}{4}$ and $-4$.
The lines will be $y=\dfrac{x}{4}+c$ and $y=-4x+k$ respectively. each line passes through $\left( 1,5 \right)$.
Putting the point, we get the constants as $c=5-\dfrac{1}{4}=\dfrac{19}{4}$ and $k=5+4\times 1=9$ respectively.
Therefore, the lines are $y=\dfrac{x}{4}+\dfrac{19}{4}\Rightarrow 4y=x+19$ and $y=-4x+9\Rightarrow 4x+y=9$.

Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance and hence we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to $\infty$.