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Given: Parallelogram MNPQ, M(3a-b,b) N(3a+b,3b) P(6a+4b,3a+3b)

Find: The quordiantes of point Q

Dec 3, 2018

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Given: Parallelogram MNPQ, M(3a-b,b) N(3a+b,3b) P(6a+4b,3a+3b)

Find: The coordinates of point Q

The opposite sides are parallel

So....the slope between  MN is

[ ( 3b - b) ]                           2b

_________________  =   _____

[ (3a + b) - ( 3a - b)]            2b

And the slope joining PQ is the same....let (x,y) be the coordinates of Q

[ (3a + 3b)  - y]                      2b

__________________ =   ______

[ (6a + 4b) - x ]                      2b

This implies that  x = (6a + 2b)  and y = (3a + b)

So Q =   [ 6a + 2b , 3a + b ]

We can check this..since NP and MQ are parallel.....the slope of NP should = the slope of MQ

NP

[ (3a + 3b) - 3b]                    3a                    a

_________________ =     _______  =      ____

[(6a + 4b) - (3a + b)]           3(a + b)            a + b

MQ

[ b - ( 3a + b) ]                          - 3a         -3a                      a

_________________  =        _____ =   ________   =    _____

[ (3a - b) - (6a +2b) ]             -3a -3b      -3 ( a + b)          a + b

Dec 4, 2018