Question

The points (2, –1), (3, 4), (–2, 3) and (–3, –2) taken in the order from the vertices of _________________

• A. Square
• B. Rectangle
• C. Rhombus
• D. Triangle

Answer 1

The correct option is C Rhombus

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

Distance $A{B}^2=(3-2{)}^2+(4+1{)}^2$
= 1 + 25
= 26
AB = √26

Distance $B{C}^2=(-2-3{)}^2+(3-4{)}^2$
= 25 + 1
= 26
BC = √26

Distance $C{D}^2=(-3+2{)}^2+(-2-3{)}^2$
= 1 + 25
= 26
CD = √26

Distance $D{A}^2=(-3-2{)}^2+(-2+1{)}^2$
= 25 + 1
= 26
DA = √26

Since all four side are equal, it can be a square or rhombus.
Lets check the diagonals

Distance $A{C}^2=(-2-2{)}^2+(3+1{)}^2$
= 16 + 16
= 32
AC = √32

Distance $B{D}^2=(-3-3{)}^2+(-2-4{)}^2$
= 36 + 36
= 72
CD = √72

Since the diagonals are not equal we can see that the figure is a rhombus