+20 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).
+526 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.
