Does a linear equation with two variables when plotted on a graph always give a straight line?

Question

I have read that linear equations with two variables when plotted always give a straight line. In wikipedia,I read that linear equation is that a linear polynomial equated to zero. We know that linear polynomial has degree or highest variable power 1.So 3x+4 or 3x+4y^0 can be a example of linear polynomial. But if we plot 3x+4y^0=0 then we get a straight line but when y is 0 x is undefined........

So it is not a straight line as it is discontinous.Please clear my doubt.Hope you understand my question .I don't know much as I am student ...Please clear the concept.

Answer 1

In general, we say that something like $$ x^2 + 3x + 2 $$ is a polynomial of degree 2. It's "nice" to write this as $$ x^2 + 3 x^1 + 2 x^0 $$ because there's a clear pattern: the powers of $x$ decrease to zero.

But the use of $x^0$ here should be regarded as "formal" -- it's just a different way of writing the number $1$, but which fits into the nice pattern. So what you've asked about: $$ 3x^1 + 4 y^0 $$ looks like it's be undefined at $y = 0$, but in fact it's really a peculiar way of writing $$ 3x^1 + 4 \cdot 1, $$ i.e., $$ 3x + 4 $$ which is defined for every $xy$-point in the plane.

By asking this, you've hit on a particularly messy bit of notation, one that we (as mathematicians) tend to gloss over because we all "know what it means." To actually formally define what a polynomial is takes a bit of work, and what we teach beginning algebra students is just a way of getting them accustomed to working with polynomials without all the formal stuff...but then it runs into nasty cases like the one you've raised.

So: good for you for looking closely at what you're working with!

But: ignore this problem, and write your polynomials without any $x^0$ or $y^0$ terms, and use constants instead. Then the most general linear polynomial you need to consider is $$ Ax + By + C $$ where $A, B,$ and $C$ are all real numbers, and at least one of $A$ or $B$ is nonzero. (If they were both zero, then this would be a degree-0 polynomial.)

If you set this to be zero, you get $$ Ax + By + C = 0 $$ and if $B$ is nonzero, you can rewrite that as $$ y = -\frac{A}{B} x - \frac{C}{B} $$ which is a line in $y = mx + b$ form. If $B$ actually IS zero, then you have $$ Ax + C = 0\\ x = \frac{-C}{A} $$ which is the equation of the vertical line through the point $(-\frac{C}{A}, 0)$.