If A(–2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

OR

If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.(CBSE 2018)

Given:

Parallelogram ABCD

A( – 2, 1), B(a, 0), C(4, b), D(1, 2)

Now, Midpoint of AC = Midpoint of BD

(∵ diagonals of a parallelogram bisect each other)

Using Midpoint Formula i.e.

The midpoint of the segment joining and is given by

Putting values,

The midpoint of AC

The midpoint of BD 
Now,
a + 1 = 2
a = 1
And
1 + b = 1
b = 0

⇒ a = 1 and b = 0

Using Distance Formula i.e.

Distance between (x₁, y₁) and (x₂, y₂ ) is given by:

D = 

Putting Values,

AB = √(0 + 9) = 3 = CD

AD = √(36 + 16) = √52 = 2√13 = BC

So,

AB = CD = 3; AD = BC = 2√13

OR

Ar(ABCD) = Ar(ABC) + Ar(ADC)

The area of triangle with points (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is:

• Ar(ABC)

• Ar(ADC)

So, Ar(ABCD)