Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-

coordinate geometry problem

Name: Anonymous 2008-11-26 8:37

In coordinate geometry:
Given the midpoints of the sides of a triangle: P(2,5), Q(4,2) and R(1,1)
Find the coordinates of the vertices.

And tell me how you solved it.

Name: Anonymous 2008-11-26 9:12

Find the line parallel to PQ that passes through R, then the same for the others points and lastly find the intersections of the lines.

Name: Anonymous 2008-11-27 8:01

HURR DURR.
if they are midpoint of triangle, then the resultant triangle from the midpoints would bisect perpendicularly to the original triangle. go from there fag00.

Name: Anonymous 2008-11-27 9:21

just do it with a straightedge on graph paper.

Name: Anonymous 2008-11-27 15:10

>>4

How about no!

Name: Anonymous 2008-11-27 18:29

Midpoint between 2 points (X1,Y1) and (X2,Y2) is ((X1+X2)/2,(Y1+Y2)/2).
I'd make 3 points (X1,Y1),(X2,Y2),(X3,Y3) and then you can create a system of equations that you can easily solve.

Name: Anonymous 2008-11-30 20:13

Greetings anon, on this wonderful evening. >>1 through >>6 are completely Phail. To solve this, one needs to simply use vectors and the parallelogram rule, i.e., let a,b and c represent the vertices of the larger triangle.

Thus, b = \stackrel{\rightarrow}{RP} + \stackrel{\rightarrow}{RQ}, c = \stackrel{\rightarrow}{PR} + \stackrel{\rightarrow}{PQ} and a = \stackrel{\rightarrow}{QR} + \stackrel{\rightarrow}{QP}.

Name: Anonymous 2008-11-30 20:18

Fuck me.

b = \vec{RP} + \vec{RQ}, c = \vec{PR} + \vec{PQ} and a = \vec{QP} + \vec{QR}.

Name: Anonymous 2008-12-03 11:25

>>7
>>8

How about you fail and they were right!
No one said nothing about vectors.

Name: Anonymous 2008-12-04 20:24

Thus, Unknown control sequence '\stackrel', Unknown control sequence '\stackrel' and Unknown control sequence '\stackrel'.

Don't change these.
Name: Email:
Entire Thread Thread List