Forming the statement mathematically(And ignoring the constant C since we're taking only the x and y coordinates):
Let $y = mx$ be our line
Let $(x_1, y_1)$ and $(x_2, y_2)$ be the two points that satisfy the equation.
Hence, we can write the statements as:
$y_1 = m\cdot x_1$ and $y_2 = m\cdot x_2$
Now, how do we prove that:
The equation is satisfied for the following:
$(ax_1+bx_2, ky_1+ly_2)$ i.e.
$(ky_1+ly_2) = m(ax_1+bx_2)$
where a, b, k and l are constants.
This is definitely not true in general. In other words, you have too many degrees of freedom.
Counterexample: $m=2,$ and $(x_1,y_1)=(1,2), \; (x_2,y_2)=(2,4).$ You can verify that both points are on the line $y=2x.$ Now let $a=1,\; b=2,\; k=3, \; l=4$. By your claim, the point $(1\cdot 1+2\cdot 2, 3\cdot 2+4\cdot 4)=(5, 22)$ should be on the line, but it certainly is not.
What is true is that any linear combination of the points $(x_1,y_1)$ and $(x_2,y_2)$ will also be on the line. So let's take $2(1,2)+5(2,4)=(12, 24).$ You can see at a glance this is on the line.
In comparing what is true with what is not, I conclude that you need to modify your claim to say that $\mathbf{a}=\mathbf{k}$ and $\mathbf{b}=\mathbf{l}.$ That is, you must multiply both the $x$ and $y$ coordinates of your points by the same value.