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					                                2x+y=3




x+2y=3                       (0,3)




                                     (1.5,0)

         (-3,0)

                  (-1.5,0)

                                         (3,-3)
                         B

                   h

                   R
                        A         O

                                            C




                                         D
Diagonals AC and BD divide the parallelogram into four triangles of equal
area. Therefore, Area of ΔAOB= (1/4)Area of ||gm ABCD=2sq. Units
Area of ΔAOB=1/2*AO*h⇒4=2*h⇒h=2
2x+y=3 m1=-2
x+2y=3 m2=-1/2               Tanθ=|m1-m2|/|1+ m1m2|=3/4
h/OB=3/5 ⇒ OB=10/3 ⇒BD=20/3
h/OR=3/4 ⇒ OR=8/3 ⇒AR=8/3-2=2/3
AB2=AR2+RB2=(4/9)+4=40/9 ⇒AB=2√10/3
RC=4+2/3=14/3
BC2=RB2+RC2 ⇒BC2=(196/9)+4 ⇒BC2=232/9=2.2.2.29/9 ⇒BC=2√58/3
Hence AB=2√10/3, BD=20/3 and BC=2√58/3

				
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