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

To do: To find the values of a and b, Hence find the lengths of its sides.

Solution:

M is midpoint of AC and BD

Using midpoint formula,

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

$\Rightarrow\frac{2}{2}=\frac{a+1}{2}$  and  $\frac{b+1}{2}=\frac{2}{2}$

$\Rightarrow a+1=2$ and $b+1=2$

$\Rightarrow a=2-1$ and $b=2-1$

$\Rightarrow a=1$ and $b=1$

we know  if there two points $( x_{1}, y_{1})$ and $( x_{2}, y_{2})$,

Distance between the two points,$=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Using the distance formula,

Length of its sides,

$AB=BC=CD=DA=\sqrt{(-2-a)^{2}+(1-0)^{2}}$

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

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

$\Rightarrow AB=\sqrt{9+1}$

$\Rightarrow AB=\sqrt{10}$ unit

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

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

$\Rightarrow BC=\sqrt{10}$ unit 

Thus $a=1, b=1$ and length of its sides$=AB=BC=CD=DA=\sqrt{10}$ unit.

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Updated on: 10-Oct-2022

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