# General equation in $\mathbb{R^2}$

In my notes, the following is written: The most general equation of a line in $\mathbb{R^2}$ is $ax+by=c,$ Where $a,b,c \in \mathbb{R}$ and $a,b$ not both $=0.$

I understand this, is it just something trivial, and a simple demonstration of a linear equation?

My question(s) is/are:

Are all linear equations in $\mathbb{R^2}$ simply straight lines, and linear equations in $\mathbb{R^3}$ represent planes?

And suppose, in the example above, $a=b=0$. Would $c=0$ represent a point in the one dimensional plane?

Just an idea, probably wrong $\ddot\smile.$

• You mean "the general equation for a line", not "the most general equation". – quasi Mar 16 '17 at 22:42
• @quasi my lecturer wrote 'the' most general equation...', but I would understand it as the standard/general equation of a line? – Gurjinder Mar 16 '17 at 22:45
• For your other question, recall that for an equation with at most two vaiables $x,y$, the graph of the equation, by definition, is the set of all points $(x,y)$ which satisfy the equation. Hence if $a = b = c = 0$, the graph is ...? And if $a=b=0$ but $c \ne 0$, the graph is ...? – quasi Mar 16 '17 at 22:46
• Well, if you consider a system of equations it gets more general, and that would allow you to transform the space so to speak. But perhaps this is the only simple equation. Does that help? – theREALyumdub Mar 16 '17 at 22:46
• @theREALyumdub mmm, not really, sorry, I'm quite new to the introduction of mathematical rigour and goemetric intuitions. What would you mean by 'transforming the space's? – Gurjinder Mar 16 '17 at 22:58

Are all linear equations in $\mathbb R^2$ simply straight lines? Yes, in the sense that you specified above (nonzero $a$ and $b$). The reason is this: The equation can be rewritten (in my opinion more intuitively) as $$ax+by=c \quad \Leftrightarrow \quad \Biggl\langle\begin{bmatrix}a\\b\end{bmatrix} \Bigg| \begin{bmatrix}x\\y\end{bmatrix} \Biggr\rangle = c$$ where $\langle \boldsymbol u| \boldsymbol v\rangle$ is the scalar product $\boldsymbol u^\top \boldsymbol v$. The right equation should always evoke the intuition of the variable point $[x,y]^\top$ being projected onto the constant reference $[a,b]^\top$. That is the key feature of the scalar product. If $\|[a,b]^\top\|=1$, then $c$ is exactly the length of that part in $[x,y]^\top$ that is parallel to $[a,b]^\top$. If $\|[a,b]\| = A \neq 1$, then $c/A$ is the parallel length.
Either way, the scalar product expression essentially says: "I hold for all points $[x,y]$ that have a constant parallel part with $[a,b]$." If you draw it on a piece of paper, that means that $[a,b]^\top$ is perpendicular to the line of all possible $[x,y]^\top$, which intersects $[a,b]^\top$ at a distance of $c/A$ (from the origin $[0,0]^\top$).
Do all linear equations linear equations in $\mathbb R^3$ represent planes? Yes, in the same sense as above. Consider: $$ax+by+cz=d \quad \Leftrightarrow \quad \Biggl\langle\begin{bmatrix}a\\b\\c\end{bmatrix} \Biggl| \begin{bmatrix}x\\y\\z\end{bmatrix} \Biggr\rangle = d$$ which holds for all points $[x,y,z]^\top$ that have with $[a,b,c]^\top$ a parallel part of $d/\|[a,b,c]^\top\|$, and an arbitrary perpendicular part. Again, $[a,b,c]$ specifies the normal direction, and the "arbitrary perpendicular part" has two degrees of freedom, meaning: All points on a plane that is offset by $d/\|[a,b,c]^\top\|$ in direction $[a,b,c]^\top$.
In the example above, $a=b=0$. Would $c=0$ represent a point in the one dimensional plane? Nope, but we don't even need geometric intuitions for this one. Just look at the resulting equation: $$0\cdot x + 0 \cdot y = 0$$ For which choice of $x$ and $y$ is it satisfied? The answer will give you the set of $[x,y]^\top$ that this equation defines. Do they all lie on the same line? Or does this particular equation provide a different set of $[x,y]^\top$ points altogether?