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# Vectors and bisects? watch

1. I don't understand this example in the Edexcel M1 textbook so any help would really be appreciated -

OABC is a parallelogram. P is the point where the diagonals OB and AC intersect. The vectors a and c are equal to OA--> and OC--> respectively. Prove that the diagonals bisect each other.

Their working is as follows, and the bit I don't understand is in italics:

'From the diagram,
OB = OC + CB = c + a
and AC = AO + OC = -OA + OC = - a + c

P lies on OB, therefore OP = u( c - a ) (where u is a co-efficient)P lies on AC, therefore OP = a + p( -a + c) (where p is a co-efficient)

a + p( -a + c) = u( c - a )
....'

They then go on to equate coefficients to prove that the lines bisect as P is the midpoint of both.

I don't get why they added in the a to the equation to find P on AC - why didn't they simply use the equation they formed for AC with a co-efficient? Like they did for the other simultaneous equation?
2. (Original post by yhnah)

I don't understand this example in the Edexcel M1 textbook so any help would really be appreciated -

OABC is a parallelogram. P is the point where the diagonals OB and AC intersect. The vectors a and c are equal to OA--> and OC--> respectively. Prove that the diagonals bisect each other.

Their working is as follows, and the bit I don't understand is in italics:

'From the diagram,
OB = OC + CB = c + a
and AC = AO + OC = -OA + OC = - a + c

P lies on OB, therefore OP = u( c - a ) (where u is a co-efficient)P lies on AC, therefore OP = a + p( -a + c) (where p is a co-efficient)

a + p( -a + c) = u( c - a )
....'

They then go on to equate coefficients to prove that the lines bisect as P is the midpoint of both.

I don't get why they added in the a to the equation to find P on AC - why didn't they simply use the equation they formed for AC with a co-efficient? Like they did for the other simultaneous equation?
OP is parallel to OB so you can say that OP = t OB for some constant t.

But OP is not parallel to AC so you can't say that OP = t AC for some constant t.

Instead, OP = OA + AP = a + t AC.
3. (Original post by notnek)
OP is parallel to OB so you can say that OP = t OB for some constant t.

But OP is not parallel to AC so you can't say that OP = t AC for some constant t.

Instead, OP = OA + AP = a + t AC.
Thank you! I get it now, that should have been obvious

Thanks again!

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Updated: February 11, 2016
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