How to prove that linear transformations transforms lines into lines

The most general way to define a line is to specify a point on the line and a direction that the line runs parallel to. This way, the line can be written: $$L(t) = \vec{p} + \vec{d}t$$ where $\vec{p}$ is any point on the line, $\vec{d}$ is the direction it runs in, and $t$ is a real number parameter. Allowing $t$ to run from negative infinity to positive infinity gives all points on the line.

To show a linear transformation turns straight lines into straight lines, apply it to $L(t)$ and see if the result can be put into a similar form.

Lines are parallel if they have the same direction vector. If a transform preserves parallelness (parallelity? parallelocity? parallelitude? paralexity? I'll stop now.) then the resultant lines should have the same direction vector.

One thing that is important to note, is because we're dealing with a linear transformation, we're not dealing with an affine space and we only have to consider "lines" that intersect the origin. In this case, that means that if you have two parallel lines, they actually have the same "slope-intersect form", which you may remember from high school math. Because they're the same and $T$ is obviously well-defined, it's going to send them to the same new "line", which is about as parallel as the previous line was with itself.

Consider the form of the linear equation,

$$ y = lx + m $$

Well, we know $m = 0$, so we have:

$$ y = lx $$

Furthermore,

$$ T(x,y) = T(x,lx) $$

From here, we already know that both terms of our image will be expressible with one variable, $x$, so we know that the second term will be expressible as $\hat{l}x $ for some constant $\hat{l}$, and we'll have a linear equation and it is a "straight line". We can go ahead and compute what $\hat{l} $ is anyways.

$$ T(x,lx) = (ax + blx, cx + dly) = (x(a+bl),x(c+dl)) $$

Now we can just take out tuple and re-express it as an equation:

$$ f(x(a+bl)) = x(c+dl) \Leftrightarrow f(x_1) = x_1\frac{c+dl}{a+bl} $$

$$ \Rightarrow \hat{l} = \frac{c+dl}{a+bl} $$

And we have our new equation,

$$ y_1 = \hat{l}x_1 $$

Which parameterizes a fantastically straight line.

What your teacher was probably trying to do, was be a little clever and ask you to prove the mapping $T$ is a linear transformation in the first place.