G C S E

[kuiperbelt]

Please see link below:

https://ibb.co/6ByjcM5

[ZebuSanctuary]

To avoid syntax errors the following stipulation will apply;

→
AB will be written as AB

alternately

→
BA = BA

The direction of displacement arrow will be omitted.

AB = AW + WO + OB ----> -b + -2b + a

AB = a - 3b

Since X is the midpoint of AB then

AX = AB/2 ----> (a - 3b)/2

AX = a/2 - 3b/2

Now find WX in terms of a & b

WX = WA + AX ----> b + a/2 - 3b/2

WX = a/2 - b/2 or (a - b)/2

Note the scalar quantity 1/2 will be expressed as 0.5, thus

Vector WX = 0.5(a - b)

Determine WY in terms of a & b

WY = WO + OB + BY ----> -2b + a + a (BY = a as point B is midpoint along OY)

WY = 2a - 2b

where 2 is a scalar quantity;

Vector WY = 2(a - b)

Finally find XY

XY = XA + AW + WO + OB + BY alternately use XY = XB + BY

XY = -a/2 + 3b/2 - b - 2b + a + a

XY = 3a/2 - 3b/2 common factor 3/2 is scalar and will be represented as 1.5 hence

Vector XY = 1.5(a - b)

It can be seen from the 3 vectors WX, WY and XY the coefficients of a and b are +1 and -1 respectively. It follows that points W, X and Y are colinear and sit on the straight line WXY.