Question

Show that the points (4,6),(-1,5)(-2,0) and (3,1) are the vertices of a rhombus.

Answer 1

Answer:

Consider the given points.

A(0,0),B(3,4),C(0,8),D(−3,4)

Distance of AB,

AB=

(3−0)

2

+(4−0)

2

AB=

(3)

2

+(4)

2

AB=

9+16

=

25

=5

Similarly,

BC=

(0−3)

2

+(8−4)

2

BC=

(−3)

2

+(4)

2

BC=

9+16

=5

Similarly,

CD=

(−3−0)

2

+(4−8)

2

CD=

(−3)

2

+(−4)

2

CD=

9+16

=5

Similarly,

DA=

(0+3)

2

+(0−4)

2

DA=

(3)

2

+(−4)

2

DA=

9+16

=5

Therefore,

AB=BC=CD=AD

Now,

AC=

(0−0)

2

+(8−0)

2

AC=

(0)

2

+(8)

2

AC=

64

=8

Similarly,

BD=

(−3−0)

2

+(4−8)

2

BD=

(−3)

2

+(−4)

2

BD=

9+16

=5

So,

AC



=BD

As all the sides are equal and diagonals are not equal.

Its shows that the following vertices are of rhombus.

Hence, this is the answer.

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