Question: A person considers lines on the plane $\mathbb{R^2}$ to be solutions of equations of the form $ax+by+c=0$, where $a,$ $b,$ and $c$ are fixed reals satisfying $a^2+b^2\neq0$. Give a point $P=(x_0,y_0)$ show an equation of a line passing through $P$ and parallel to the line given by $ax+by+c=0$.
My work so far:
Lines that are parallel have the same slope. So, if I put $ax+by+c=0$ into slope intercept form, I end up with $y=\frac{-ax}{b}-\frac{c}{b}$. So my slope is $m=\frac{-a}{b}$. Using this information, I will find the y-intercept using the point $P$. So I get: $b=y_0-(\frac{-a}{b})x_0$ $\Rightarrow$ $b=y_0+\frac{a}{b}x_0$. Now, putting all this information together, back into the slope intercept form I have: $y=\frac{-a}{b}x+y_0-(\frac{-a}{b})x_0$.
I'm not sure if any of this is correct or not. Any help would be appreciated.
All lines parallel to $ax+by+c=0$ are of the form $ax+by+d=0$ for some $d$.
If the parallel line passes through $(x_0, y_0)$, then $ax_0+by_0+d=0$ or $d=-ax_0-by_0$.
The line is therefore $ax+by-ax_0-by_0=0$.
Note: I know this is substantially equivalent to some of the other answers, but I wanted to express it more simply.