Slope of a vertical line Find the equation of the line that satisfies the given conditions:
Through $(-1,2)$; parallel to the line $x=5$
I know that the equation  of this line is $x=-1$ because $x=5$ is a vertical line and any line parallel to it would be vertical as well. So the line runs vertically along $x=-1$. 
But how would I solve this algebraically? 
 A: The reasoning you gave in your question is algebraic: I see nothing wrong with it. You can't use the standard "parallel lines have equal slopes" since your lines have undefined slopes.

However, here is another approach. All lines can be put in general linear form:
$$Ax+By=E$$
You first find $A$, $B$, and $E$ (non-unique) for your starting line $x=5$. You want to find another line,
$$Cx+Dy=F$$
that is parallel to your line, which happens when
$$AD-BC=0$$
Find $C$ and $D$ that fulfill that requirement, and find $F$ so that the point $(-1,2)$ is on the line. Then you are done!

If you want still another, slightly different approach, you could use the form
$$x \cdot \cos \theta + y \cdot \sin \theta = r$$
where $\theta$ is the direction angle of a vector perpendicular to the line and $r$ is the distance of the origin from the line. Two lines are parallel if their $\theta$'s are equal or differ by a multiple of $\pi$. In your particular problem, $\theta$ is zero.
A: Consider the line$$x=5$$
$$0=x-5$$
$$0y=x-5$$
$$y=\frac{x}{0}-\frac{5}{0}$$
$$y=\infty x-5\infty$$
Now try to determine the slope of a line perpendicular to this line.
$$m_\perp = -\frac{1}{m}=-\frac{1}{\infty}=0$$
Then try to find a line that passes through $(-1,2)$ with a slope of $m_\perp=0$
$$m_\perp=0=\frac{y-2}{x+1}$$
$$0(x+1)=y-2$$
$$y-2=0$$
$$y=2$$
$$y=0x+2$$
Now consider a line that passes through (-1,2) of the form 
$$y=ax+b$$
where $a$ is non-zero. The slope of a line that is perpendicular to this line has a slope of $-1/a\ne 0$, therefore, this line cannot be perpendicular to $y=0x+2$. If $a=0$, then the 2 lines coincide (and a line is parallel to itself, and therefore not perpendicular to itself). So we can conclude that the line is not of this form $y=ax+b$.
The only other form of line is modeled with the equation
$$x=b$$
substituting we have
$$-1 = b$$
and so, the line is
$$x=-1$$
