linear equation for vertical line

Question

If we want to graph a horizontal line, we will do the following:

y = 0x + 3

No matter the domain for x, the range for y will always be 3. Therefore, we have a horizontal line.

y = 0(0) + 3 = (0,3)
y = 0(1) + 3 = (1,3)
y = 0(2) + 3 = (2,3)

Now the formula to graph a vertical line looks like this:

x = 3

Well, wait a second. Where is the y? I would like to see the y in the equation. But it is missing. How can I write the equation for a vertical line that includes the y variable? This is all I can think of:

x = 0y + 3

And with the following domain:

x = 0(0) + 3
x = 0(1) + 3
x = 0(2) + 3

Is this correct? Is it ok to reverse the x and y, as I just did above? Or does this not make it a slope-intercept equation anymore? It should still be a linear equation, since the variables are raised to the first power, in my opinion. But the slope-intercept form looks like this: y = mx + b. So I am not sure if this is still a slope-intercept equation.

Answer 1

No matter which form you choose your equation to a unique line will always be unique, For example $$\frac{x}{a}+\frac{y}{b}=1$$ or $$Ax+By=C$$ or $$y=mx+c$$ or $$m=\frac{y-y_0}{x-x_0}$$or$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$or$$x=x_0+at, y=y_0+bt$$ Whichever form you choose it'll always lead you to a unique solution (Just sometime it'll not look similar but believe me it'll be, It's just matter of rearrangement )

Answer 2

For linear functions we have the ability to write one variable solely in terms of the other, with $y$ being the usual dependent variable. The expression $y=2x+5$ is shorthand for the set $\{(x,y)\in \mathbf R^2\mid y = 2x+5\}$, points whose $y$ coordinate is five more than twice the $x$ coordinate; at the end of the day you're just specifying points whose coordinates meet a certain criteria. The equation $x=3$ simply indicates the set of points $\{(x,y)\in \mathbf R^2\mid x=3\}$, points whose $x$ coordinate is 3.

Also, since a linear function with nonzero slope is 1-1, it is possible to write $x$ as a function of $y$: $$ y = mx+b\quad\Rightarrow \quad mx = y-b \quad\Rightarrow \quad x = \frac1m y - \frac bm$$