Question

prove that the points (2, -1), (0, 2), (3, 3) and (5, 0) are the vertices of a parallelogram. Also find the angle between its diagonals.

Answer 1

Hence, it is parallelogram proved.

Step-by-step explanation:

Let the points A(2, -1), B(0, 2), C(3, 3) and D(5, 0) are the vertices of a parallelogram.

∴ $AB^{2}$ = $(0 - 2)^{2}$ + $(2 + 1)^{2}$

= $( - 2)^{2}$ + $(3)^{2}$

= 4 + 9 = 13

$BC^{2}$ = $(3 - 0)^{2}$ + $(3 - 2)^{2}$

= $3^{2}$ + $1^{2}$

=  9 + 1 = 10

$CD^{2}$ = $(5 - 3)^{2}$ + $(0 - 3)^{2}$

= $(2)^{2}$ + $( - 3)^{2}$

= 4 + 9 = 13

$DA^{2}$ = $(5 - 2)^{2}$ + $(0 + 1)^{2}$

= $(3)^{2}$ + $(1)^{2}$

= 9 + 1 = 10

Diagonas,

$AC^{2}$ = $(3 - 2)^{2}$ + $(3 + 1)^{2}$

= $(1)^{2}$ + $(4)^{2}$

= 1 + 16 = 17

$BD^{2}$ = $(5 - 0)^{2}$ + $(0 - 2)^{2}$

= $(5)^{2}$ + $(- 2)^{2}$

= 25 + 4 = 29

∴ $AB^{2}$ = $CD^{2}$ = 13 and

$BC^{2}$ = $DA^{2}$ = 10

And

$AC^{2}$≠ $BC^{2}$

Hence, it is parallelogram.