While the previous answers correctly made it clear that straight lines are not was defines a linear transformation, it is actually possible to visualise more general linear transformations with such lines. However, you need to come clear about where these lines lie: it has nothing to do with the trigonometric functions' graphs. After all, these functions are just special elements of the domain of the Fourier transform.
What matters is the graph of the transform itself. As was already said, the Fourier transform maps functions to functions, which is considerably more complicated than mapping numbers to numbers. However, functions form a vector space: naïvely, you can write a tuple something like
$$
\begin{pmatrix} \vdots \\ f(0) \\ f(0.000\ldots0001) \\ f(0.000\ldots0002) \\ \vdots \\ f(1) \\ \vdots \end{pmatrix}
$$
of all the function values. Strictly this space is infinite-dimensional, but in many applications you're only interested in some finite range and resolution anyway, making the space finite-dimensional. Still rather too big to visualise, but for the extreme case $n=2$ it's actually almost possible: our "functions" $f(x)$ can now be seen as points in a plane with coordinate $(f(x_1),f(x_2)) =: (f_0,f_1)$. The Fourier transform reduces to just two different modes as well: "in phase" and "anti-phase" or "frequency $0$" and "frequency $1$", i.e.
$$\begin{aligned}
F_0 &= \left(\sqrt{\tfrac12},\sqrt{\tfrac12}\right) \\
F_1 &= \left(-\sqrt{\tfrac12},\sqrt{\tfrac12}\right)
\end{aligned}$$
That shows already a matrix representation of the (inverse) Fourier transform, and a matrix representation is enough to show that the transformation is linear. But you still don't see the lines, all right...
The graph of this Fourier-transform function (which maps now $\mathbb{R}^2\to\mathbb{R}^2$) lives in $\mathbb{R}^4$. Still a bit bulky that space, but let's look a 3-dimensional subspace: we put in only values where the coefficient of $F_1$ is zero, i.e. the part of the graph we look at consists of triples
$$
\begin{pmatrix} a_0 \\ a_0\cdot(F_0)_0 + 0 \\ a_0\cdot(F_0)_1 + 0 \end{pmatrix}
= \begin{pmatrix} a_0 \\ \sqrt{\tfrac12}a_0 \\ \sqrt{\tfrac12}a_0 \end{pmatrix}
$$
and if we follow the path of these points take through $\mathbb{R}^3$ while varying $a_0$ we see – exactly: a straight line, leading away from the origin in an oblique angle!
The actual graph is not a line, but a plane in $\mathbb{R}^4$.