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Graph the points A(-5, 0), B(3, 2), C(5, 6), and D(-3, 4).

Prove that ABCD is a parallelogram by showing that the

diagonals bisect each other (have the same midpoint).

 Apr 21, 2021
 #1
avatar+275 
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The points are A(-5,0), B(3,2), C(5,6), D(-3,4) and diagonals are BD and AC. 

Let midpoint of BD be (x,y).

 

\(x = {3-3 \over 2}\)

   \(x = 0\)

\(y = {2+4 \over 2}\)

    \(y = 3\)

∴ Midpoint of BC = (0,3)

 

Let midpoint of AD be (x',y')

 

\(x' = {-5+5 \over 2}\)

   \(x' = 0\)

\(y' = {0+6 \over 2}\)

    \(y' = 3\)

∴ Midpoint of AD = (0,3) 

 

Thus, diagonals AC and BD bisect each other.

Hence ABCD is a ||gm. 

 Apr 21, 2021
 #2
avatar+20 
+1

Thank You so much!! amygdaleon305

 Apr 21, 2021

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