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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.

mathmaniage
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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.

Adrian Keister
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  • I'm confused by your third paragraph. Any linear combination of two points is a sum of scalar multiples of the two points. The claim is true if and only if the vectors represented by the two points are linearly dependent. Otherwise their linear span will be whole of $\mathbb{R}^2$ – user160738 Jul 13 '18 at 17:47
  • @user160738: You wrote, "... if the vectors represented by the two points are linearly dependent." That's the same thing as saying that they both lie on the same line through the origin, which is exactly what the OP stipulated must hold. Essentially, all that's being said here is that all the points on a given line through the origin form a vector space. So, in my answer, the assumptions the OP made should be assumed in my answer as well. – Adrian Keister Jul 13 '18 at 17:51