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URGENT HELP HOMEWORK DUE TOMORROW

Posted January 4th, 2010 by isuckatgeomertry
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Okay the question is....
The midpoints of the sides of a triangle are (3,2), (2,4), and (3,5).
What are the coordinates of the vertices

‹ THIS IS ALSO URGENT! PLEASE HELP! Straight line ›

Comments

answer

On January 4th, 2010 Structure says:

This is very similar with this one:

Find out vertices of an triangle from its midpoints

I've adapted it with your numbers:

A1 = (3,2) is the middle of BC
B1 = (2,4) is the middle of CA
C1 = (3,5) is the middle of AB

Let $ A = (x_a,y_a), B = (x_b,y_b), C=(x_c,y_c) $ be the desired points.
We know that:
B+C=2A1
C+A=2B1
A+B=2C1
so
$ (x_b,y_b)+(x_c,y_c)=2(3,2)=(6,4) $
$ (x_a,y_a)+(x_c,y_c)=2(2,4)=(4,8) $
$ (x_a,y_a)+(x_a,y_a)=2(3,5)=(6,10) $

Summing up these three equalities we have
$ 2(x_a+x_b+x_c,y_a+y_b+y_c)=(16,22)=2(8,11) $
$ (x_a+x_b+x_c,y_a+y_b+y_c)=(8,11) $

Then
$ A=(x_a,y_a)=(x_a+x_b+x_c,y_a+y_b+y_c)-(x_b,y_b)-(x_c,y_c)= $
$ =(8,11)-(6,4)=(2,7) $

$ B=(x_b,y_b)=(x_a+x_b+x_c,y_a+y_b+y_c)-(x_a,y_a)-(x_c,y_c)= $
$ =(8,11)-(4,8)=(4,3) $

$ C=(x_c,y_c)=(x_a+x_b+x_c,y_a+y_b+y_c)-(x_a,y_a)-(x_b,y_b)= $
$ =(8,11)-(6,10)=(2,1) $

thank you so much will you

On January 4th, 2010 isuckatgeomertry says:

thank you so much will you please look at my other post and help me with that. Thankss

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