Question

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

Answer 1

We know that diagonals of a parallelogram bisect each other.

Coordinates of the midpoint of AC = coordinates of the midpoint of BD 

the midpoint of AC = midpoint of BD

${\displaystyle \Rightarrow (\frac{4-2}{2},\frac{b+1}{2})=(\frac{a+1}{2},\frac{0+2}{2})}$

${\displaystyle \Rightarrow (\frac{2}{2},\frac{b+1}{2}) =(\frac{a+1}{2},\frac{2}{2})}$

${\displaystyle \Rightarrow (1,\frac{b+1}{2})=(\frac{a+1}{2},1)}$

So

${\displaystyle 1=\frac{a+1}{2}}$`

2 = a + 1

${\displaystyle \therefore a=1}$

and  

${\displaystyle \frac{b+1}{2}=1}$

${\displaystyle \Rightarrow b+1=2}$

${\displaystyle \therefore b=1}$

Therefore, the coordinates are A(–2, 1), B(1, 0), C(4, 1) and D(1, 2).

${\displaystyle AB=DC=\sqrt{{(1+2)}^2+{(0-1)}^2}=\sqrt{9+1}=\sqrt{10}}$

${\displaystyle AD=BC=\sqrt{{(1+2)}^2+{(2-1)}^2}=\sqrt{9+1}=\sqrt{10}}$