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Prove 3 points are collinear

I was wondering if you guys could help me with this question.

The question is:

The points A, B and C have position vector of a, b and c respectively referred to an origin O.

a) Given the point X lies on AB produced so that AB:BX=2:1, find x, the position vector of X, in terms of a and b

b) If Y lies on BC, between B and C so that BY:YC=1:3, find y, the positoin vector of Y, in terms of b and c.

c) Given that Z is the mid-point of AC, show that X, Y, and Z are collinear

d) Calculate XY:YZ

My answers:

a) x=3/2b-1/2a

b) y=1/3c+3/4b

c) I don't know how to do this one, since I don't know how to get them to be scalar multiples. By the way, this is what I think the question is like. I'm guessing I don't understand the question cause I don't see how the points could be collinear.

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Thanks in advance

Re: Prove 3 points are collinear

Verasi

Answer a is not correct.

consider 3 position vectors OA , OB and OC ,so that B lies on the line AC and the ratio (AB)/(BC) = m/n where m and n are two real numbers.

then the vector OB =[(nOA+mOC)/(m+n)

Apply this formula : in your case m=2 n=1 OC=OX =X (unknown) , OB =b and OA =a : then b = [2x+1a]/3 make x the subject and find x = [3b-a]/2

this is the correct answer that gives the position vector OX in terms of a and b.

work and check the other answers......

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Re: Prove 3 points are collinear

I agree with your answer to part a). I think you made an arithmetic error for part b); it should be Y=3/4B +1/4C. The theorem of Menelaus shows X, Y and Z are collinear. I don't know how to answer part d) without a little linear algebra. If you know some, a change of basis to X and Y from B and C yields Z = -X + 2Y. I've attached a diagram: Oops, I just noticed I didn't answer part d): YZ, the distance from Y to Z, is |-X + 2Y -Y| = |Y - X| the distance from X to Y. So YZ:XY = 1:1.

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Re: Prove 3 points are collinear

Thanks for the answer, but can you give me a link where I can find the proof of the form of Menelaus Theorem used in the problem or give me a proof? Thanks again. Also, what does x' y' and z' stand for?

Re: Prove 3 points are collinear

Hi,

If P = (a,b), Q = (c,d) are coordinates of points P and Q and x is a real number. xP = (xa, xb) and P + Q = (a+c, b+d). Then a point xP + x'Q is on the line through P and Q iff x + x' =1. This is a standard way of treating points algebraically; if you like, the set of points in the plane form a vector space of dimension 2.

The only reference that I know off the top of my head is an old (but still very good book) on geometry by Dan Pedoe.

Namely, Geometry, A Comprehensive Course. This form of Menelaus is on page 28. I didn't check the site cut-the-knot, but he might very well have this form.