Show that the following points taken in order form the vertices of a parallelogram. ( - 7, - 5), ( - 4, 3), (5, 6) and (2, –2)

Answers (1)

Let A,B,C and D represent the points (-7,-5),(-4,3),(5,6) and(2,-2)
Distance of AB
AB=sqrt(-4+7)^2+(3+5)^2
AB=sqrt3^2+8^2
AB=sqrt9+64
AB=sqrt73

Similarly 
Distance of CD
CD=sqrt(2-5)^2+(-2-6)^2
CD=sqrt(-3)^2+(-8)^2
CD=sqrt(9+64)
CD=sqrt73

Similarly 
Distance of DA
DA=sqrt(-7-2)^2+(-5+2)^2
DA=sqrt(-9)^2+(-3)^2
DA=sqrt81+9
DA=sqrt90

Similarly
Distance of BC
BC=sqrt(5+4)^2+(6-3)^2
BC=sqrt9^2+3^2
BC=sqrt81+9
BC=sqrt90

So,AB=CD=sqrt73 and BC=DA=sqrt90 
i.e., The opposite sides are equal
Hence ABCD is a parallelogram.

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