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

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

$x=\left(\frac{x_1+x_2}{2}\right)andy=\left(\frac{y_1+y_2}{2}\right)$

$O$ $(\frac{-2+4}{2},\frac{1+b}{2})$ ..................(1)

Mid-point let say $P$ of diagonal $BD$ is given by

$P$ $(\frac{a+1}{2},\frac{0+2}{2})$ ..................(2)

Points $O$ and $P$ are same

Equating the corresponding co-ordinates of both midpoints, we get

$\frac{-2+4}{2}=\frac{a+1}{2}$

$\Rightarrow a=1$

and

$\frac{1+b}{2}=\frac{0+2}{2}$

$\Rightarrow b=1$

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, $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

we can find the length of each side

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

$AB=\sqrt{(3{)}^2+(1{)}^2}=\sqrt{10}$

$AB=CD$ ...............(pair of opposite sides of the parallelogram are parallel and equal)

$BC=\sqrt{(4-1{)}^2+(1-0{)}^2}$

$BC=\sqrt{(3{)}^2+(1{)}^2}=\sqrt{10}$

$BC=AD$ ...............(pair of opposite sides of the parallelogram are parallel and equal )

$\Rightarrow AB=BC=CD=AD=\sqrt{10}$

$\Rightarrow ABCD$ is a $Rhombus$