Question

Show That the quadrilateral whose vertices at the point ( 2 ',-1 ) , ( 3 , 4 ) , ( -2,3 ) and (-3,-2 ) is a rhombus .

Answer 1

AB=BC=CD=AD=5unit
AD||BC,AB||CD reason is that all sides are equal

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

P(2, -1), Q(3, 4) ,R(-2, 3) and S(-3 , -2)

Now,

Using Distance Formula,

PQ = √(x2 - x1)² + (y2 - y1)²

PQ = √(3 - 2)² + (4 - 1)²

PQ = √(1)² + (5)²

PQ = √(1 + 25)

PQ = √26 units

Also,

QR = √(-2 - 3)² + (3 - 4)²

QR = √(-5)² + (-1)²

QR = √25 + 1

QR = √26units

Also,

RS = √(-3 + 2)² + (-2 - 3)²

RS = √(-1)² + (-5)²

RS = √1 + 25

RS = √26 units

Also,

SP = √(- 3 - 2)² + (- 2 + 1)²

SP = √(-5)² + (-1)²

SP = √25 + 1

SP = √26 units

Hence we get,

All the sides are equal in length

Therefore

It is proved that it is a rhombus.