Question

If$A(-2,1),B(9,0),C(4,6)$ and $D(1,2)$ are the vertices of $ABCD$. Hence find the length of all side of quadrilateral $ABCD$.

Answer 1

In a parallelogram the diagonals bisect each other

Midpoint of AC = midpoint of BD

Midpoint formula$\frac{(x1+x2)}{2};\frac{(y1+y2)}{2}$

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

$\frac{2}{2};\frac{(1+b)}{2}=\frac{(a+1)}{2};\frac{2}{2}$

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

$a+1=2$

$a=1$

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

$1+b=2$

$b=1$

$a=1andb=1$

Using distance formula we came find out the length of the sides of parallelogram

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

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

$=\sqrt{(9+1)}$

$=\sqrt{10}$

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

$=\sqrt{3^2+1^2}$

$=\sqrt{10}$

It is a rhombus as all the sides are equal