+0  
 
0
73
1
avatar

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
 #1
avatar+98196 
+1

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 

 

 

 

cool cool cool

 Dec 4, 2018

14 Online Users

avatar
avatar