1. ## Algebraic Vectors

Describe how you could use vectors to determine if 3 given points are collinear. Test your method using each set of points.

$\displaystyle i) P(-3, 1), Q(2, 4), R(5,6)$

$\displaystyle ii) D(5,1), E(1, -5), F(-3,-11)$

I know how to find 2 points if they are collinear or not, but I don't know how to find 3 points that are collinear or not. . .so how would you determine them?

I know. . .

$\displaystyle \frac {x_1}{x_2} = \frac {y_1}{y_2}$

2. Originally Posted by Macleef
Describe how you could use vectors to determine if 3 given points are collinear. Test your method using each set of points.

$\displaystyle i) P(-3, 1), Q(2, 4), R(5,6)$

$\displaystyle ii) D(5,1), E(1, -5), F(-3,-11)$

I know how to find 2 points if they are collinear or not, but I don't know how to find 3 points that are collinear or not. . .so how would you determine them?

I know. . .

$\displaystyle \frac {x_1}{x_2} = \frac {y_1}{y_2}$
use 2 points to set up the equation of a straight line and test afterwards if the 3rd point is located on the line:

$\displaystyle PQ: (x,y) = (-3, 1) + r (5, 3)$ . If R(5, 6) lies on the line the coordinates of R must satuisfy the equation of the line:

$\displaystyle (5, 6) {\buildrel ? \over = }(-3,1) + r(5,3)$ . That means from

$\displaystyle 5 = -3 + 5r ~\implies ~ r = \frac85$ and from

$\displaystyle 6 = 1 + 3r~\implies~ r = \frac53$ there doesn't exist one value for r so that the point R is "produced" by the equation. Thus $\displaystyle R \notin PQ$

I'll leave the next problem for you.

(For confirmation only: F is placed on DE)