# proving sum of any scalar multiple of two points lie on the line passing through them

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.

• What are a, b, k, and l? – Adrian Keister Jul 13 '18 at 14:03
• This is true iff the two points and the origin are collinear. You need to specify that proviso. – Somos Jul 13 '18 at 14:38
• @AdrianKeister, they are just scalar constants. – mathmaniage Jul 13 '18 at 16:07
• @Somos, it is actually true for any points, I've verified experimentally for some points, even for fractions, even when they are not collinear to the origin, and I have mentioned that I avoided the constant 'c' because the only points of concern are x and y. – mathmaniage Jul 13 '18 at 16:09

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.
• 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