Find the equation of a line which is parallel to the line mx + ny + c = 0 and passing through the point (2,3).
The given answer is: m(x-2)+n(y-3) = 0, but how do I get there?
The lines $\displaystyle mx+ny+c=0$ for fixed $\displaystyle n$ and $\displaystyle m$ form a series of parallels as $\displaystyle c$ varies.
So the line you seek is found by finding the value of $\displaystyle c$ so that $\displaystyle mx+ny+c=0$ goes through $\displaystyle (2,3)$. That is:
$\displaystyle 2m+3n+c=0$
or:
$\displaystyle c=-(2m+3n)$
CB
Any line parallel to $\displaystyle Mx+Ny+C=0$ has the same form $\displaystyle Mx+Ny+D=0$.
If it contains the point $\displaystyle (2,3)$ this must be true:
$\displaystyle 2M+3N+D=0$ or $\displaystyle D=-(2M+3N)$
$\displaystyle Mx+Ny-(2M+3N)=0$
$\displaystyle M(x-2)+N(y-3)=0$