Determine the equation of a line

[Ana.stasia]

The question is:
Determine the equation of a line that contains a point (1/3, 2/3) and is parallel to the ordinate axis.

This is how I tried to do it. I got 1/3. The result says it should be x=1/3. Although it's similar the whole process seems wrong. Especially the 2/3=1/3 part.

 

[lev888]

The question is:
Determine the equation of a line that contains a point (1/3, 2/3) and is parallel to the ordinate axis.

This is how I tried to do it. I got 1/3. The result says it should be x=1/3. Although it's similar the whole process seems wrong. Especially the 2/3=1/3 part.

The problem is that vertical lines do not have the slope intercept form equations. Why? Their slope is undefined. Rise over run is not defined if run is 0.
The equation is x=1/3. Meaning a point lies on this line as long as its x coordinate is 1/3.

 

[Ana.stasia]

The problem is that vertical lines do not have the slope intercept form equations. Why? Their slope is undefined. Rise over run is not defined if run is 0.
The equation is x=1/3. Meaning a point lies on this line as long as its x coordinate is 1/3.

I am not sure I understand. So they don't have m or n?

 

[JeffM]

I am not sure I understand. So they don't have m or n?

No. The only relevant value for x is 1/3 because x = 1/3. And y can take on any value.

 

[Dr.Peterson]

I am not sure I understand. So they don't have m or n?

A vertical line has only an x-intercept (your "m"). It never intersects the y-axis, so it can't have a y-intercept (your "n").

Also, its slope (gradient) is undefined. (In America, it is traditional to use m for the slope.)

 

[Ana.stasia]

A vertical line has only an x-intercept (your "m"). It never intersects the y-axis, so it can't have a y-intercept (your "n").

Also, its slope (gradient) is undefined. (In America, it is traditional to use m for the slope.)

Okay. So because I don't have an "n" it's just y=kx, and k is undefined. So that brings us to:
Y=undefined*x
Now, why is the equation in the result just x=1/3?

 

[lev888]

I am not sure I understand. So they don't have m or n?

You can't represent a vertical line by a function y=kx+n. Why? Because a vertical line is NOT a function. A function has one y value for any x. A vertical line has an infinite number of y values for one x value.

 

[Mr. Bland]

You can't have a vertical line as a function [MATH]y = f(x)[/MATH], but I wouldn't say it can't be represented as a function. Could flip it around and use [MATH]x = my + b[/MATH] or something a bit more accommodating like [MATH]ax + by + c = 0[/MATH].

The takeaway, though, is that if the slope is given as the ratio of "rise over run", a vertical slope has a run of zero and therefore an undefined division results.

 

[Dr.Peterson]

I wouldn't say it can't be represented as a function.

It can't be represented as a function of x. The general form ax+by+c=0 isn't a function; but as you say, it can be a function of y, or it could be a parametric function. Neither of those is in view here.

Okay. So because I don't have an "n" it's just y=kx, and k is undefined. So that brings us to:
Y=undefined*x
Now, why is the equation in the result just x=1/3?

No, it can't be y = kx, because k, like n, is not defined. More generally, that form would be a line through the origin at slope k; it implies that your m and n are both 0.

Don't they teach you these things? Commonly, when you learn the "slope-intercept form", you are also taught that it applies to any line that is not vertical, and you are taught the forms y=n (horizontal) and x=m (vertical) as special cases. Here, you need the latter. You might want to search your textbook for such information.