# Formula y=kx+n help

1. Mar 28, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

Can somebody use the formula y=kx+n, to present me formula for the line which is parallel to the y axis and go among the point (-4,-5)?

2. Relevant equations

y=kx+n

3. The attempt at a solution

k=tg(alpha)

k=tg90

k=$$\frac{1}{0}$$

y=$$\frac{x}{0}$$

x=0

-4=0

Hm.....

2. Mar 28, 2008

### Feldoh

If the line is parallel to the Y AXIS then the slope has to be undefined, see why?

Last edited: Mar 28, 2008
3. Mar 28, 2008

### rock.freak667

you can't multiply by zero like that.

The y-axis is the vertical one,right?

4. Mar 28, 2008

### sutupidmath

wouldn't that be just x=-4?

5. Mar 28, 2008

### rock.freak667

That's what I believe it will be.

6. Mar 29, 2008

### Physicsissuef

I know that it will be x=-4. But please prove it with y=kx+n
rock.freak667, y is the vertical one.

7. Mar 29, 2008

### kamerling

The problem is that the two assumptions

k, and n exist such that {(x,y): y=kx+n} is the set of points of the line

{(x,y): x = -4} is the set of points of the line

can not both be true.

8. Mar 29, 2008

### Physicsissuef

what is k, what is n?

9. Mar 29, 2008

### HallsofIvy

You can't. A vertical line, parallel to the y-axis, cannot be written in that form.

10. Mar 29, 2008

### Physicsissuef

Why?

11. Mar 29, 2008

### HallsofIvy

For exactly the reason everyone has been telling you! Writing "y= kx+ n" means that different values of x give different values of y. That is true for all lines except vertical lines. A line parallel to the y-axis has the same value of y for every x. When you learned "y= kx+n", the "slope-intercept" form, you should have learned that every non-vertical line can be written in that form but vertical lines cannot.

Last edited by a moderator: Mar 29, 2008
12. Mar 29, 2008

### Physicsissuef

Ahhh... I undertand now. Thank you very much.
btw- what is that same value for y?

13. Mar 29, 2008

### HallsofIvy

My mistake. I meant to say the "same x value for every y". In this case, since you tell us one point is (-4, -5) that x value is x= -4 for all y. That's why we can write the equation of that line "x= -4" without mentioning y.

(A horizontal line would have the "same y value for every x". The horizontal line (parallel to the x-axis) passing through (-4, -5) would have y= -5 for every y. Of course, that is of the form "y= kx+ n" with k= 0, n= -5.)