91Ó°ÊÓ

A quadrilateral with vertices $(0,0),(1,3),(4,2)$ and$(3,-1)$

Show that the quadrilateral is a square.

For this, show that lines making adjacent sides of quadrilateral are perpendicular and all sides are equal.

Slope, $m=\frac{y_2-y_1}{x_2-x_1}$

Slope of line joining vertices $(0,0)$and $(3,-1)$is: $m_1=\frac{-1-0}{3-0}$

$m_1=-\frac{1}{3}$

Slope of line joining vertices $(3,-1)$and $(4,2)$is: $m_2=\frac{2-(-1)}{4-3}$

$m_2=3$

Slope of line joining vertices $(4,2)$and $(1,3)$is: $m_3=\frac{3-2}{1-4}$

$m_3=-\frac{1}{3}$

Slope of line joining vertices $(1,3)$and $(0,0)$is: $m_4=\frac{0-3}{0-1}$

$m_4=3$

As $m_1×m_2=-\frac{1}{3}×3=-1$

$m_2×m_3=3×-\frac{1}{3}=-1$

$m_3×m_4=-\frac{1}{3}×3=-1$

$m_4×m_1=3×-\frac{1}{3}=-1$

So, the adjacent sides of the quadrilateral are all perpendicular to each other.

Distance formula of line joining two points $(x_1,y_1)$and $(x_2,y_2)$is $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Length of side joining vertices $(0,0)$and $(3,-1)$is: $\sqrt{{(3-0)}^2+{(-1-0)}^2}=\sqrt{9+1}=\sqrt{10}$

Length of side joining vertices $(3,-1)$and $(4,2)$is: localid="1646113403237" $\sqrt{{(4-3)}^2+{(2-(-1\left)\right)}^2}=\sqrt{1+9}=\sqrt{10}$

Length of side joining vertices localid="1646113785453" $(4,2)$and localid="1646113798928" $(1,3)$is:localid="1646113476115" $\sqrt{{(1-4)}^2+{(3-2)}^2}=\sqrt{9+1}=\sqrt{10}$

Length of side joining vertices localid="1646113817275" $(1,3)$and localid="1646113823270" $(0,0)$is: localid="1646113557192" $\sqrt{{(0-1)}^2+{(0-3)}^2}=\sqrt{1+9}=\sqrt{10}$

All sides of the quadrilateral are equal.