Show that the points A(3, 0), B(4, 5), C(– 1, 4) and D(– 2, – 1) are the vertices of a rhombus. Find its area.
AB = √{(4 – 3)2 + (5 – 0)2} = √{1 + 25} = √26 units
BC = √{(– 1 – 4)2 + (4 – 5)2} = √{25 + 1} = √26 units
CD = √{(– 2 – (– 1))2 + (– 1 – 4)2} = √{1 + 25} = √26 units
DA = √{(– 2 – 3)2 + (0 – 1)2} = √{25 + 1} = √26 units
AC = √{ (– 1 – 3)2 + (4 – 0)2} = √{32}
BD = √{(– 2 – 4)2 + (– 1 – 5)2} = √{36 + 36} = 6√2units
Since AB = BC = CD = DA
Hence, ABCD is a rhombus
Area = 1/2 × (product of diagonals)
= 1/2 × 4√2 × 6√2
= 24 sq units
28