1. ## Points on line?

I want to know if there's a way to find out if several points form a straight line. I could of course use y=kx+m but i don't wanna solve it that way. I could check if the points have the same y-value but different x-values. But how do I check if the line is a diagonal?
e.g let's say I have theese points: P1(2,1), P2(4,3), P3(6,5), P4(8,7). I can easily visualize 'em and see that they form a straight line, but how can I solve it without drawing them?

2. $\displaystyle \vec{P_1P_3},\vec{P_1P_4}$ are proportional vectors to $\displaystyle \vec{P_1P_2}$

Fernando Revilla

3. Ok, and then I have to do gauss-jordan Elimination? Hm... seems there's no easy way around this Thanks Fernando

4. Originally Posted by crimshaft
Ok, and then I have to do gauss-jordan Elimination? Hm... seems there's no easy way around this

You needn't Gauss Jordan Elimination and you needn't Lax Milgran Theorem either .

Fernando Revilla

5. Originally Posted by FernandoRevilla
You needn't Gauss Jordan Elimination and you needn't Lax Milgran Theorem either .

Fernando Revilla
What do you propose? How do I otherwise check if they're proportional to eachother? What method should I read about?

6. Originally Posted by crimshaft
e.g let's say I have theese points: P1(2,1), P2(4,3), P3(6,5), P4(8,7). I can easily visualize 'em and see that they form a straight line, but how can I solve it without drawing them?
$\displaystyle P_2-P_1=<2,2>~\&~P_2-P_3=<-2,-2>$ because those are multiples of each other $\displaystyle P_1,~P_2,~\&~P_3$ are collinear.

7. Ah ok. But i'm searching for a way to know if the points sum up to a straight line that is if they all lie on the same line. If I used that solution wouldn't it indicate that two paralell lines lie on the same line?

8. Originally Posted by crimshaft
What do you propose? How do I otherwise check if they're proportional to eachother? What method should I read about?

$\displaystyle (a,b)$ is proportional to $\displaystyle (c,d)\neq (0,0)$ iff $\displaystyle ad-bc=0$ .

Fernando Revilla

9. Sry, my bad. I get it now. Though I might use the one point formula in the end. Think it will lead to a faster algorithm. Thanks!

10. Similar but perhaps a little faster: If all the points lie on a single line, the slopes will be the same:
$\displaystyle \frac{y_1- y_0}{x_1- x_0}= \frac{y_2- y_0}{x_2- x_0}= \frac{y_3- y_0}{x_3- x_0}= \cdot\cdot\cdot= \frac{y_n- y_0}{x_n- x_0}$