Question

If $A(-2,1)$, $B(a,0)$, $C(4,b)$ and $D(1,2)$, are the vertices of parallelogram $ABCD$, find $a$ and $b$. Hence find the length of it's sides.

Answer 1

In a $\parallel gm$ diagonals bisect each other, therefore mid-points of diagonals are equal.

Mid - point of AC= Mid-point of BD

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

$1,\left(\frac{1+b}{2}\right)=\left(\frac{a+1}{2}\right),1$

$\left(\frac{a+1}{2}\right)=1\Rightarrow a=2-1=1$

$\left(\frac{b+1}{2}\right)=1\Rightarrow b=2-1=1$

Distance $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

$=\sqrt{10}$

Distance of $BC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

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

$=\sqrt{10}$

Hence, it is a rhombus, therefore

$AB=BC=CD=AD$