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Prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other, using vectors.
I tried this,
but failed.
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hi Freiza,
Welcome to the forum.
I have embedded your image into your post so that members will not have to go to that site to view it.
Your method should work but it looks to me like there are many errors.
Let's try to sort them out one at a time. I'll leave out the vector over-lining to save time.
You want the vector for QN.
I would work like this:
QN = QM + MN = QA + AM + MB + BN = d/2 + a/2 + a/2 + b/2 = (d + 2a + b)/2
Are you able to work out PM in a similar way? Please post back.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Already tried that,
PM = QM - QP
QM = (a+d)/2
QP = - (c+d)/2
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hi Freiza,
Ok, that looks better now.
But the intersection has to be relative to the origin:
and
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Three things I found out today,
1) I need to learn more.
2) You are Intelligent, your parallelogram trick was good.
3) You were born on a Monday
Thank You very much.
But why MN is not relative to origin, ie why it is not (a+d)/2 + (a+b)/2, i know this is not correct but I am failing the intuition here.
Last edited by Freiza (2017-02-22 00:03:42)
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Freiza,
I shall use your notation for proof.
Last edited by thickhead (2017-02-22 14:06:50)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Thank You thickhead,
But for future reader a little correction, QH = (3a +2b + c + 2d)/ 4
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3) You were born on a Monday
I think you're confusing me with Solomon Grundy
But why MN is not relative to origin, ie why it is not (a+d)/2 + (a+b)/2, i know this is not correct but I am failing the intuition here.
The method works by finding the equation of the two lines and then setting them equal to find the point of intersection.
To construct such an equation:
Step (1) Start at the origin.
Step (2) Write a vector from the origin to anywhere on the line.
Step (3) Go an unknown amount in the direction of the line using a parameter.
One of the lines goes through the origin so step (2) is not required.
In coordinate geometry it is usual to use x for the independent variable and y for the dependent variable.
In vector geometry a Greek letter and a vector 'r' are usually used.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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feeiza,
I use capital letters for the point location and small letters for position vector. that always gives a disciplined approach rather than using free vectors.the origin acts as anchor so that you won't go astray.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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