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.

Given sides of parallelogram A(-2,1), B(a,0), C(4,b), D(1,2)
We know that diagonals of Parallelogram bisect each other.
Mid-point let say O of diagonal AC is given by x=(x1+x2)/2 and y=(y1+y2)/2
O(-2+4/2, 1+b/2). (1)
Mid-point let say P of diagonal BD is given by P(a+1/2, 0+2/2). (2)
Points O and P are same
Equating the corresponding co-ordinates of both midpoints, we get
-2+4/2 = a+1/2 => a=1
and 1+b/2 = 0+2/2 => b=1
Now the Given co-ordinates of the parallelogram are written as A(-2,1), B(1,0), C(4,1), D(1,2)
By distance formula, √(x2-x1)²+(y2-y1)² we can find the length of each side
AB=√(-2-1)²+(1-0)²
AB=√(-3)²+(1)²=√10
AB=CD.. (pair of opposite sides of the parallelogram are parallel and equal)
BC=√(4-1)²+(1-0)²
BC=√(3)²+(1)²=√10
BC=AD.. (pair of opposite sides of the parallelogram are parallel and equal )
=>AB=BC=CD=AD=√10
=>ABCD is a Rhombus