Question

Show that the following points taken in order form the vertices of a rhombus.( - 4, - 7), ( - 1, 2), (8, 5) and (5, - 4)

Answer 1

A( - 4, - 7), B( - 1, 2), C(8, 5), D(5, - 4)
XY= √(x2- x1)² + (y2- y1)² )
AB=√ ((-1-(-4))²+ (2-(-7))²
= √(3²+ 9²)
=√(90)
AB= 3√10
BC= √(8-(-1))²+ (5-2)²
  = √9²+3²
BC= 3√10
CD= √(5-8)²+ (-4-5)²
CD= 3√10
AD= √(5-(-4))²+(-4-(-7))²
AD = 3√10
∴ AB=BC=CD=AD
AC=√(8-(-4))²+ (5-(-7))²
     =√144+144
AC=12√2
BD=√(5-(-1))²+(-4-2)²
    = √36+36
BD= 6√2
∴AC≠BD
As all the sides are equal and diagonals are not equal
its shows that the following vertices are of rhombus

Answer 2

Answer:

First (i): All sides are equal so its an Square

Second (ii): All sides are different so its an trapezium

Third (iii): AB=CD and BC=DA or AD

so two parallel sides are equal

its an parallelogram

HOPE YOU GET IT..!  

Step-by-step explanation: