Find a line that is perpendicular from x=10 and one point, (4,5) The title is quite self-explainary ;)
I know that the answer is $y=5$, but I am not sure how does one person come to that conclusions showing all the steps. Can you please help me here?
 A: You could do this in many ways.
1) It is obvious for some.

2) You can visualize this on your head, or even plot this graph and see clearly.

3) This is the only method that I could think at this time of which you could present if you were asked to prove it.
$x= 10$ is a vertical line, A perpendicular line to this must be parallel to the x axis. Hence it must have a gradient 0. So it is in the form $y= (0)x+ c=c$ 
We know two points on this line we are finding which are : 
$(4,5)$ and $(10,b)$ 

Where b is an unknown 

So let's apply the formula for the gradient : 
$$\frac{5-b}{4-10}=0$$
$$b=5$$
Now we know the line is in the form $$y=c$$ and since $y=5$ (of any of the two points) we can conclude that the line is $$y=5$$
A: I had to do my own version, using @Tharindu's knowledge for my assignment. Here it is if anyone wants a more detailed version. The coordinates are different, since I had a different question:
We need to find an equation that will create a line that is perpendicular to the equation $x=3.58$, given that one point on the line B. As shown in steps beforehand, the point $B$ is equal to $(14.99,8.28)$ 
Any equation in the form $x=c$ will always make a vertical line. A perpendicular line is a line that intersects on a right angle. Therefore, knowing that $x=3.58$ will always be a vertical line, the perpendicular line will a horizontal line. A horizontal line always has a $m$ value of 0. 
Therefore, using the base equation for a linear equation: 
$$y=mx+c$$
Sub in the known values :
$$y=(0)x+c$$
Simplify 
$$y=c$$
That is the equation of our horizontal line as of now. Now, we know one point on this horizontal line, $B$, and we know the $m$ value of the equation is 0. 
With this information, we can find the equation of the line using the point-slope formula: 
$$y−y_1=m(x−x_1)$$
Where $y_1$ and $x_1$ can be any point on the line. In this case, we will use the $B$ coordinate.
Co-ordinate of B: $(14.99,8.28)$
Let x=14.99 be $x_1$
Let y=8.28 be $y_1$ 
$(14.99,8.28)$=$(x_1,y_1)$
Point-Slope Formula: 
$$y−y1=m(x−x1)$$
Sub in known values: 
$$y−y1=m(x−x1)$$
$$y−8.28=m(x−14.99)$$
$$y−8.28=0$$
$$y=8.28$$
Therefore, the equation that is perpendicular to the directrix, given that one point on that line has the coordinate (14.99,8.28), is y=8.28
