HOW TO PROVE THE GIVEN FOUR POINTS FORM A PARALLELOGRAM USING SLOPE

Example 1 :

Using the concept of slope, show that the vertices (1 , 2), (-2 , 2), (-4 , -3) and (-1, -3) taken in order form a parallelogram.

Solution :

Let the given points be A(1 , 2) B(-2 , 2) C (-4 , -3) and D (-1, -3)

Slope of AB  =  (y2 - y1)/(x2 - x1)

A(1 , 2) B(-2 , 2) 

  =  (2 - 2) / (-4 - 1)

  =  0/(-5)  =  0  ---(1)

Slope of BC  =  (y2 - y1)/(x2 - x1)

B(-2 , 2) and C (-4 , -3) 

  =  (-3 - 2) / (-4 + 2)

  =  -5/(-2)  =  5/2  ---(2)

Slope of CD  =  (y2 - y1)/(x2 - x1)

C (-4 , -3) and D (-1, -3)

  =  (-3 + 3) / (-1 + 4)

  =  0/5  =  0 ---(3)

D (-1, -3) and A(1, 2)

  =  (2 + 3) / (1 + 1)

  =  5/2---(4)

(1)  =  (3) and (2)  =  (4)

Hence the given points form a parallelogram.

Example 2 :

Show that the opposite sides of a quadrilateral with vertices A(-2 ,-4), B(5 , -1), C(6 , 4) and D(-1, 1) taken in order are parallel.  

Solution :

Slope of AB  =  (y2 - y1) / (x2 - x1)

A(-2 ,-4), B(5 , -1)

  =  (-1 + 4) / (5 + 2)

  =  3/7  ----(1)

Slope of BC  =  (y2 - y1) / (x2 - x1)

B(5 , -1) and C(6 , 4) 

  =  (4 + 1) / (6 - 5)

  = 5/1

=  5 ----(2)

Slope of CD  =  (y2 - y1) / (x2 - x1)

 C(6 , 4) and D(-1, 1)

  =  (1 - 4) / (-1 - 6)

  = -3/(-7)

=  3/7 ----(3)

D(-1, 1) and A(-2, -4)

  =  (-4 - 1) / (-2 + 1)

  = -5/(-1)

=  5 ----(4)

(1)  =  (3)

(2)  =  (4)

Hence the opposite sides in a quadrilateral are parallel.

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