Two equal representations of a line on a plane The Definition of line is $L=v+w\mathbb{R}$ where $w\neq 0$. 
I want to understand why this Definition of a line is equal to this Definition:
A line (in $\mathbb{R^2}$) is defined as $L=\{(x,y)\in\mathbb{R}^2:ax+by=c\}$ where $(a,b)\neq(0,0)$
There is a proof in my book but I don't understand a part of it I hope somebody can help.
If $a=0$, then $b\neq 0$ the equation is therefore
$$by=c\quad \text{so}\quad y=\frac{c}{b}$$
Setting $v=(0,\frac{c}{b})$ and $w=(1,0)$ then $L=v+\mathbb{R}w$
I understand that if we pick an element from the set $v+\mathbb{R}w$ then it has the form $(\lambda,\frac{c}{b})$ and this is a solution to the equation $ax+bx=c$. I don't understand it however for the opposite case that if we take a solution $(x,y)$ it must be in $v+\mathbb{R}w$
I have the same Problem when it cames to the case where $a$ is not equal to $0$
Is $a\neq0$ then settig $y=0$ we have a solution $(x,y)$ for the equation $ax+bx=c$ where $x$ is $\frac{c}{a}$ and $y=0$
We set this solution $(\frac{c}{a},0)$ to be $v$
Setting $c=0$ helps to find $w$
We have $L_0=\{(x,y)\in\mathbb{R}^2:ax+by=0\}$
Then $0\in L_0$ Setting $y=1$ gives 
$$w=(-\frac{b}{a},1)\in L_0$$ 
Now Picking an element $z$ of the set $v+\mathbb{R}w$
The elemeent $z$ has the form $(\frac{c-\lambda b}{a},\lambda)$ which is a solution for the equation $ax+by=c$
Conversely I don't know why if I pick a solution to the equation $ax+by=c$ it also must be in $v+\mathbb{R}w$
Can someone explain it to me please?
 A: All of this only holds in $\mathbb R^2$. Let's see how we get from $L=v+w\mathbb R$ to $ax+by=c$. Take any vector $n=(a,b)$ such that $\langle n,u\rangle =0$. Then, $L'=n\mathbb R$ defines a line which is orthogonal to $L$. If $(x,y)\in L$ is a point, $(x,y)-v$ is a vector pointing in the same direction as $w$, i.e. a multiple of $w$. You can easily see this by writing $(x,y)=v+tw$ for some $t\in\mathbb R$. But then, $\langle (x,y)-v,n\rangle=0$, since $n$ is orthogonal to $w$. We can in fact write $L$ as the set of points $(x,y)$ such that $(x,y)-v$ is orthogonal to $L$. You should draw a picture to see this geometrically!
Now, $$L=\{(x,y)\in \mathbb T^2\mid \langle (x,y)-v,n\rangle=0\}$$ and since $n=(a,b)$, this gives $$0=\langle(x,y)-v,n\rangle=\langle(x,y),(a,b)\rangle-\langle v,n\rangle=ax+by-c,$$ where $c=\langle v,n\rangle$ and this is how one goes from the parametric form to the equation.
If you are given an equation $ax+by+c=0$ for $L$, you can get the parametric form by writing $(x,y)=(v_1,v_2)+t(w_1,w_2)$ for unknowns $v_1,v_2,w_1,w_2$ and solving a linear system of equations.
