I have a line, parallel to the $y$-axis, that passes through a point, P:
$$P(1/2,-3/5)$$
What is the equation of the line?
What I tried:
$$(y−y_0)=m(x−x_0)$$
$$(y+3/5)=m(x−1/2)$$
Though, there exists no $m$, therefore I can't figure out what the equation of the line is.
A line parallel to the $y$-axis has a fixed $x$-value, therefore its equation is of the form $$x=c,\quad c\in\mathbb{R}$$ where you can easily see that as you vary the value of $y$, you are drawing a line perpendicular to the $x$-axis. Since the point $(x_0,y_0)=(1/2,-3/5)$ belongs to that line, then all the points with the same $x$-value will do too. Finally you have: $$x=\dfrac{1}{2}$$
Another way to see the problem is to take a vector along the $y$-axis - for instance, the unit vector $\vec{j}=(0,1)$ - and an arbitrary vector $\vec{u}$ that defines your line, and require them to be parallel: i.e. they must be related by a constant factor.
Since you already know the coordinates of a point $(x_0,y_0)$ on your line, you can define $$\vec{u}=(x-x_0,y-y_0)=(x-1/2,y+3/5)$$ Now for the parallelism condition: $$\vec{u}=k\vec{j},\quad k\in\mathbb{R}$$ From which you get the following system of equations: \begin{cases}x-1/2=0\\y+3/5=k,\quad k\in\mathbb{R}\end{cases} The first line gives $x=1/2$, while the second tells you that $y$ can take any possible value as you vary $k$. The vector $\vec{u}$ that defines your line turns out be $\vec{u}=(0,k)$, which simply means that once you've fixed a value for $x$, then any value of $y$ is possible. And again, since $x_0=1/2$ belongs to the line, the equation of the line must be $x=1/2$.