How do you make a perpendicular line when you are given an existing line and its equation? What’s the catch?

Hint: To determine a perpendicular line when an existing line and its equation is given , we have to use the slope of the existing line for finding the slope of the perpendicular line. Then we have to use the point on the given line for determining the y-intercept of the perpendicular line.

Complete step by step solution:
If the slope of the existing line is given by $m$ then the slope of the line perpendicular to that existing line is given as follow ,
Slope of the perpendicular line , $m' = - \dfrac{1}{m}$ .
For the equation , we have to find the y-intercept of the perpendicular line which we can find with the help of the point on the existing line.
Let us understand this with an example ,
Let we have an equation of line $y = 2x + 1$ and this line passes through point $P(1, - 4)$
And let the equation of perpendicular line be $y = m^{'}x + c$
Slope of existing line is given as , $m = 2$ .
Therefore, the slope of the line perpendicular to that existing line is given as follow ,
Slope of the perpendicular line , $m' = - \dfrac{1}{m} = - \dfrac{1}{2}$ .
Now, we have to find the y-intercept of the perpendicular line ,
We have $y = m^{'}x + c$
This line passes through $P(1, - 4)$ and has slope $m' = - \dfrac{1}{2}$ .
Therefore,
$y = m^{'}x + c$
$ \Rightarrow - 4 = - \left( {\dfrac{1}{2}} \right)(1) + c$
$
   \Rightarrow - 4 = - \dfrac{1}{2} + c \\
   \Rightarrow - 4 + \dfrac{1}{2} = c \\
 $
We can write it as ,
$
   \Rightarrow c = - 4 + \dfrac{1}{2} \\
   \Rightarrow c = \dfrac{{ - 4 \times 2 + 1}}{2} \\
   \Rightarrow c = \dfrac{{ - 8 + 1}}{2} \\
   \Rightarrow c = \dfrac{{ - 7}}{2} \\
 $
We have the required equation of perpendicular line that is ,
$y = - \dfrac{1}{2}x - \dfrac{7}{2}$

Note: This type of linear equations sometimes called slope-intercept form because we can easily find the slope and the intercept of the corresponding lines. This also allows us to graph it.
We can quickly tell the slope i.e., $m$ the y-intercepts i.e., $(y,0)$ and the x-intercept i.e., $(0,y)$ .we can graph the corresponding line .