# Show an equation of a line passing through $P$ and parallel to the line given by $ax+by+c=0$.

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.

• Do you know $b \neq 0$? maybe you need to consider the $b=0$ case separately. Aug 7, 2014 at 19:13
• For an easier approach, note that the lines parallel to $ax+by+c=0$ have equations of the shape $ax+by+k=0$. Aug 7, 2014 at 19:14
• How would you go about showing that with those two equations? Aug 7, 2014 at 19:16
• Plug in $x=x_0$, $y=y_0$ and thus find $k$. Aug 7, 2014 at 19:19

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 $ax+by-ax_0-by_0=0$.
• Would you do anything with the other information provided in the problem? This is where I am confused, we have been given the information of $a^2+b^2\neq0$ and I'm not sure if we need to use this or not in our solution? Aug 7, 2014 at 20:44
• If both $a=0$ and $b=0$, then the equation $ax+by+c=0$ reduces to $c=0$, which is not the equation of a line. If $c=0$, then it's the whole plane, because any point $(x,y)$ satisfies it. If $c\neq0$, then it's an empty set of points. Including $a^2 + b^2 \neq 0$ in the question just rules out the case where both are $0$, so that all you're left with are valid linear equations. Aug 7, 2014 at 21:06