Position vectors forming a trapezium

I have a question that i'm not sure how to do.

Points A, B, C and D have position vectors a = (1,2), b = (3,5), c = (7,4) and d = (4,k)

Find one value of k for which the four points form a trapezium.

I can't remember the rules you have to use to prove it's a trapezium, and also how i would then find k. Help would be appreciated.
Original post by Ben.W_
I have a question that i'm not sure how to do.

Points A, B, C and D have position vectors a = (1,2), b = (3,5), c = (7,4) and d = (4,k)

Find one value of k for which the four points form a trapezium.

I can't remember the rules you have to use to prove it's a trapezium, and also how i would then find k. Help would be appreciated.

I'd sketch the points and determine which pair of sides are parallel. Upload what you think?
Original post by mqb2766
I'd sketch the points and determine which pair of sides are parallel. Upload what you think?

I've sketched the 3 known points and i've also drawn a dotted line where the x coordinate is 4 so i can see the line where the D coordinate lies. It looms like AB and CD will be parallel.
Original post by Ben.W_
I've sketched the 3 known points and i've also drawn a dotted line where the x coordinate is 4 so i can see the line where the D coordinate lies. It looms like AB and CD will be parallel.

Does your sketch indicate how you'll find k, given that one of the pair of sides must be parallel? Note the question asks for one value of k, so there may be more than one potential solution.
(edited 2 years ago)
Original post by mqb2766
Does your sketch indicate how you'll find k, given that one of the pair of sides must be parallel? Note the question asks for one value of k, so there may be more than one potential solution.

I'm not too sure. I remember that there's something to do with the diagonals and where they meet. Do the gradients of the parallels have to be the same maybe?
Original post by Ben.W_
I'm not too sure. I remember that there's something to do with the diagonals and where they meet. Do the gradients of the parallels have to be the same maybe?

Thats effectively what parallel means, so yes.
(edited 2 years ago)
Right, so i can work out the gradient of the parallel side of the 2 points i know and then find a value for k so that CD will have the same gradient as AB.
Original post by Ben.W_
Right, so i can work out the gradient of the parallel side of the 2 points i know and then find a value for k so that CD will have the same gradient as AB.

Yes. For completeness, note that there are two values of k which give (different) trapezium, corresponding to different pairs of sides being parallel. The question just asks for one value of k, so CD and AB being paralel is fine.
(edited 2 years ago)
Original post by mqb2766
Yes. For completeness, note that there are two values of k which give (different) trapezium, corresponding to different pairs of sides being parallel. The question just asks for one value of k, so CD and AB being paralel is fine.

Right, thank you for your help.
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