Can we have x-intercept instead of y-intercept in equation of a line

Question

As I have started learning co-ordinate geometry for ML, so got one question. Line equation

Let say in x,y dimension if a line cross over the x-axis instead of y (as shown in image above, unlike y=mx+c ) then how equation will change or will the equation has any impact apart from the x-intercept in this case? Is this a valid case? Please help me clearing this doubt.

Answer 1

One way to write an equation of an arbitrary line in the $x,y$ plane is $$ Ax + By + C = 0.$$

If $A=0$ you get a line parallel to the $x$ axis; if $B=0$ you get a line parallel to the $y$ axis.

If you set $A=m,$ $B=-1,$ and $C=b$ then the equation $Ax + By + C = 0$ describes the same line as $y = mx + b.$

But people often are interested in the equation $y = mx + b$ for reasons other than the shape it describes in a plane. We may have some quantity we can either control or observe taking different values, which we'll represent by the name $x$, and some other quantity, which we'll call $y$, whose value has some relationship to the value of $x.$

The relationship $y = mx + b$ is one of the simplest possible kinds of relationship that can occur under these circumstances. And it happens also to be possible to visualize a relationship like this by plotting a line on a graph.

A vertical line can not be the plot of such a relationship, because the first thing we wanted to see was a variety of different values of $x,$ and the vertical line has only one $x$ value. The fact that $y = mx + b$ cannot describe a vertical line therefore is irrelevant to the study of these kinds of relationship.

Answer 2

I think if the line is parallel to y - axis (as shown by the image you have attached). Then the slope of that line will not be defined because tan90 is not define .

If only slope is not define then the question of repsenting that in slope -intercept form doesn't make any sense.

Answer 3

As the other poster said, the formula basically doesn't work for vertical slopes. However, there are two ways you can save it.

The first is by rewriting it so that there is an $x$-intercept and a slope. So, if the basic formula is:

$$y = mx + b$$

We can make a similar formula for $x$:

$$x = qy + c$$

Where $q$ is the $\frac{\text{run}}{\text{rise}}$ and $c$ is the x-intercept. Here, the slope would be $q = 0$, and the $x$-intercept $c$.

Alternatively, If you wanted to "save" the original formula, you could use the hyperreals and utilize infinities. Here, you could say that the line isn't really vertical, but that its slope is indeed infinite, i.e., $\omega$ (in the hyperreals, there are more than one infinite value - you still can't "divide by zero", but you can divide by infinitely small values).

So, with hyperreals, the formula would be:

$$y = \omega x + b$$

If the $x$-intercept is $c$, then we can make the formula:

$$0 = \omega c + b \\ \omega c = -b \\ b = -\omega c $$

So, for an $x$-intercept of $c$, the new equation is

$$ y = \omega x + -\omega c$$

It's probably easier to just say to use a different formula for this situation, but I wanted to point it out because many times the places where formulas stop working because of a "divide by zero" you can often get a valid formula by extending into the hyperreals.