Answer the given question with proper steps.:

Q. 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. 

A(–2,  1), B(a, 0), C(4,  b) and D(1, 2) are the vertices of a parallelogram.

As, we know that,

Diagonals of parallelogram bisect each other.

∴ Mid point of diagonals AC = Mid point of diagonal BD

⇒ a  + 1 = 2  and  1 +  b  = 2

⇒ a  = 1 and  b  = 1

Thus, the value of

a  =1

and

b  = 1 
$$\begin{gathered} \mathrm{So} \mathrm{coordinates} \mathrm{of} \mathrm{vertex} \mathrm{are} \\ \mathrm{A}\left(-2,1\right), \mathrm{B}\left(1,0\right),\mathrm{C}\left(4,1\right) \mathrm{and} \mathrm{D}\left(1,2\right) \\ \mathrm{AB} = \sqrt{{\left(-2-1\right)}^2+{\left(1-0\right)}^2}= \sqrt{10} \mathrm{unit} \\ \mathrm{BC} = \sqrt{{\left(1-4\right)}^2+{\left(0-1\right)}^2}= \sqrt{10} \mathrm{unit} \\ \mathrm{CD}=\sqrt{{\left(4-1\right)}^2+{\left(1-2\right)}^2}= \sqrt{10} \mathrm{unit} \\ \mathrm{DA}\hspace{0.17em} = \sqrt{{\left(-2-1\right)}^2+{\left(1-2\right)}^2}= \sqrt{10} \mathrm{unit} \end{gathered}$$