Question

Show that A(-1.0). B (3, 1). C(2, 2) and
D (-2.1) are the vertices of a parallelogram
ABCD.​

Answer 1

Step-by-step explanation:

Here AB =

(3+1)

2

+(1−0)

2

=

17

BC=

(2−3)

2

+(2−1)

2

=

2

CD=

(−2−2)

2

+(1−2)

2

=

17

AD=

(−2+1)

2

+(1−0)

2

=

2

Here,AB=CD and BC=AD

Therefore its a parallelogram

Answer 2

Given :

The vertices of a parallelogram ABCD are :

$\red\bf{Values} \begin{cases} \tt{ A=(-1,0) } \\ \tt{B=(3,1) } \\ \tt {C = (2,2)} \\ \tt{ D=(-2,1)} \end{cases}$

To find :

Show that following point are the vertices of a parallelogram.

Solution :

➡ Method(1)

As we know that opposite side of parallelogram are equal.

By using Distance formula :

$\\ \implies\sf{ AB =\sqrt{ (3+1)^2+(1-0)^2} }$

$\\ \implies \sf{AB= \sqrt{17} }$

$\\ \implies\sf{ BC=\sqrt{(2-3)^2+(2-1)^2} }\\ \\ \implies\sf{BC=\sqrt{2}}$

$\\\implies\sf{CD=\sqrt{(-2-2)^2+(1-2)^2} }$

$\\ \implies\sf{CD=\sqrt{17}}$

$\\\implies\sf{AD=\sqrt{(-2+1)^2+(1-0)^2} }$

$\\ \implies\sf{AD=\sqrt{2}}$

➡ Here AB=CD and BC=AD

Therefore it's a parallelogram

____________________________________

Method (2):

By using Midpoint theorem :

$\\ \implies\sf{ Midpoint \: of \: AC =\bigg(\dfrac{-1+0}{2} ,\dfrac{0+2}{2} \bigg)}$

$\\ \implies\sf{Midpoint \: of \: AC =\bigg(\dfrac{1}{2} , 1\bigg) }$

$\\ \implies\sf{Midpoint \: of \: BD =\bigg(\dfrac{3-2}{2} , \dfrac{1+1}{2} \bigg)}$

$\\ \implies\sf{Midpoint \: of \: BD =\bigg( \dfrac{1}{2} ,1 \bigg) }$

➡ Therefore Midpoint of AC=Midpoint of BD

➡ Since diagonals of parallelogram bisect each .

Therefore ABCD is a parallelogram

Hence proved :