This is an example in the text book.

Using vectors, demonstrate that the three points A(5, -1), B(-3,4) and C(13,-6) are collinear.

Solution
AB = (-8, 5)
BC = (16, -10)
Then BC = 2AB

AB and BC have the opposite direction, so the points A, B, and C must be collinear.

I don't understand how AB = (-8, 5) and
BC = (16, -10)

vector AB is (-3 -5) i + (4 - (-1) ) j

= -8 i + 5 j

vector BC is (13 - (-3)) i + (-6 - 4) j
= 16 i - 10 j

Now the lines are collinear because they go through the same point (point B)

and they have the same slope (although one is going down the hill while the other goes up)

I'm not sur if this will help, but to get AB, you subtract A from B and B from C.

B(-3x,4y)-A(5x,-1y)=x(-3-5),y(4+1)=
AB(-8,5)

C(13x,-6y)-B(-3x,4y)=x(13+3),y(-6-4)=
BC(16,-10)

To understand how AB = (-8, 5) and BC = (16, -10), let's break it down step by step:

1. First, we need to find the vector AB, which represents the displacement from point A to point B. To calculate this, we subtract the coordinates of A from the coordinates of B.

AB = (xB - xA, yB - yA)
= (-3 - 5, 4 - (-1))
= (-8, 5)

So, AB is (-8, 5).

2. Next, we need to find the vector BC, which represents the displacement from point B to point C. We subtract the coordinates of B from the coordinates of C.

BC = (xC - xB, yC - yB)
= (13 - (-3), -6 - 4)
= (16, -10)

Therefore, BC is (16, -10).

By finding the vectors AB and BC, we can now compare them to determine the collinearity of the points. In this case, to show that the points A, B, and C are collinear, we need to show that BC is a scalar multiple of AB.

In other words, we need to show that BC = k * AB, where k is a scalar.

From the values we calculated, BC = 2AB, which means BC is equal to 2 times AB. Since BC is a scalar multiple of AB, it indicates that the points A, B, and C lie in a straight line, making them collinear.

You can verify this by observing that the vectors AB and BC have the same direction (opposite signs of the components), which further confirms the collinearity of the points.