0
$\begingroup$

I found something very mysterious for me in the slope-intercept form of the equation of the line.

First let's start with the slope equation. Assume I have 2 points (2,4) and (1,3) and I want to find slope between two points which is

$$ m = \frac{y - y_1}{x - x_1} = \frac{2 - 1}{4 - 3} = 1 $$

after we construct the slope then we continue to put it into the point-slope form:

which is $$ y-y_1 = m (x-x_1)$$ you can notice that this equation derive from the slope equation above, in the slope equation if the two points we are measuring have the same x value then the equation will be undefined. Now keep that fact in mind and move on to put value we have got into the point-slope equation:

$$ y-4 = 1 (x-2)$$ $$ y = x+ 2$$

(I use the point (2,4) in the equation) Now come to what I call mysterious part for me. if I put (2,4) in to the equation as a domain of a function then I get back $4 = 2+ 2$ this is very strange for me as a newbie because the point (2,4) was used to construct the equation which means if we reverse to the slope form we will get:

$$ 1 = \frac{y - y_1}{x - x_1} = \frac{y - 2}{x - 4} = 1 $$ which means in this case we can not put another (2,4) as a domain of the function then we would get : $$\frac{2 - 2}{4 - 4} = 1 $$ which is so wrong but why when it is in the form of $$y = x+2 $$ (which is already been construct with point (2,4)) we can use (2,4) (Again) as domain of the function?

I do not want someone to just come and show off calculation. I want to get deep to the philosophy and I want crystal clear step by step answer.

$\endgroup$
1
$\begingroup$

Your notation causes the confusion.

You have a straight line by two points, of equation

$$y=\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0.$$

Its slope is indeed

$$m=\frac{y_1-y_0}{x_1-x_0}.$$

Now you can substitute any coordinate pair $(x,y)$ into the equation.

In particular,

$$(x,y)=(x_0,y_0)\to y_0=\frac{y_1-y_0}{x_1-x_0}(x_0-x_0)+y_0,$$

and

$$(x,y)=(x_1,y_1)\to y_1=\frac{y_1-y_0}{x_1-x_0}(x_1-x_0)+y_0,$$

which are two true identities.

Your misconception comes form the fact that you are trying to change the coordinates in the slope formula, though these are fixed.


If you want to "reconstruct" the line by keeping one of the points, let $(x_0,y_0)$ and replacing the other by some $(x_2,y_2)$ drawn form the line equation, you get

$$y_2=\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0,$$ and the new equation is

$$y=\frac{\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0-y_0}{x_2-x_0}(x-x_0)+y_0,$$

which simplifies to

$$y=\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0,$$

unless $x_2=x_0$. (When this is the case, we no more have two distinct points to define the line.)

Similarly, replacing $(x_0,y_0)$ by $(x_2,y_2)$,

$$y=\frac{y_1-\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)-y_0}{x_1-x_2}(x-x_2)+\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0$$ also simplifies to

$$y=\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0$$

provided $x_1\ne x_2$, after a little more effort.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I do not understand the arrow sign, can you please change it to other form? or other sign? $\endgroup$ – user3270418 Feb 19 '16 at 14:23
  • $\begingroup$ This is an informal notation. $\endgroup$ – Yves Daoust Feb 19 '16 at 14:31
  • $\begingroup$ y= x+2 comes from (2,4) put into slope-point equation. Why when I put (2,4) again into equation. it still works? then I get 4 = 2+2 which means if I put x = 2 then I get y = 4 why the equation still work? $\endgroup$ – user3270418 Feb 19 '16 at 14:39
  • $\begingroup$ I think I fully answered. $\endgroup$ – Yves Daoust Feb 19 '16 at 14:42
  • $\begingroup$ because when I put (2,4) again into the equation. it means x_2=x_0 then I should no longer get the answer? but why I still get 2,4 as the answer? $\endgroup$ – user3270418 Feb 19 '16 at 14:43
0
$\begingroup$

The slope formula is used to find the slope of the line determined by two distinct points. You can't put the same point in twice, which is what you are trying to do--you are using $(2,4)$ and $(2,4)$. Those are the same single point, and a single point does not determine a line.

Because your line is not vertical, there is only one point on the line whose $x$-coordinate is $2$ (namely, the point $(2,4)$, of course).

When you are dealing with vertical lines, however, the slope formula can't be used. Vertical lines have undefined slope, and every point on such a line has the same $x$-coordinate.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ no, you don't get my question. the slope equation and the point-slope equation are related, when you move the (x-x_1) of the slope formula to the other side then you get slope point formula. So if you put (2,4) to construct any of one equation. it would be related $\endgroup$ – user3270418 Feb 19 '16 at 14:16
0
$\begingroup$

To begin with, you did one very confusing thing, which was you wrote $(2,4)$ and $(1,3)$ but when you substituted these into $m = \frac{y - y_1}{x - x_1}$ you wrote $\frac{2 - 1}{4 - 3}$, that is, you substituted $x$-coordinates for the $y$-variables and vice-versa. But you were inconsistent about this: in one place you wrote $ y-4 = 1 (x-2)$ but in another you wrote $\frac{y - 2}{x - 4} = 1 $.

To make things consistent, we have to rewrite some of what you wrote. Below, I chose to change the original two points to $(4,2)$ and $(3,1)$ in order to make the equation $\frac{y - 2}{x - 4} = 1 $ relevant, since that was the paradoxical equation for you.


This is not the equation of a line: $$ \frac{y - 2}{x - 4} = 1. \tag1 $$ Instead, it is the equation of a figure you get by deleting one point from a line. The point $(4,2)$ is not part of the figure.

If you add the point $(4,2)$ to that figure you get back a line which has (among other representations) the equation $$ y - 2 = x - 4. \tag2 $$

The relation between equations $(1)$ and $(2)$ is that $(1)$ implies $(2)$ (every point on the "line minus a point" is on the line) but $(2)$ does not imply $(1)$ (not every point on the line is part of the "line minus a point"--in particular, the point $(4,2)$ is in the first set but not the second set).

You can conclude $(1)$ if you are given $(2)$ and the fact that $x\neq 4$. In the case where $x = 4$ and $y = 2$, you have no justification to say that $(1)$ is true, and indeed it is not.

An equation of the form $(1)$ is indeed useful for finding the slope of a line, despite the fact that it is not completely equivalent to $(2)$. It is useful because in order to find a slope, you must have two points with two different $x$-coordinates. In effect this prevents you from writing something like $\frac{2 - 2}{4 - 4}$, in which you would have substituted the same number, $4$, for both $x$ and $x_1$ in the formula $ m = \frac{y - y_1}{x - x_1}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ after I get equation number 2 which is construct from he slope and point (4,2) why I can use the point (4,2) as the domain of the function again??? and still get the correct result y - 2 = x-4 then substitute 4,2 into equation, I get 2 - 2 = 4-4. Why??? $\endgroup$ – user3270418 Feb 19 '16 at 15:05
  • 1
    $\begingroup$ Because equation (2) is an equation of a line through the point (4,2), but equation (1) is not an equation of any line. You cannot claim that two equations should have the same solutions just because you can derive one of the equations from the other. These two equations are not equivalent. $\endgroup$ – David K Feb 19 '16 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.