Q. 114 - Precalculus Enhanced with Graphing Utilities

Question

Use slopes and the distance formula to show that the quadrilateral whose vertices are $(0,0), (1,3), (4,2)$ and $(3,-1)$ is a square.

Step-by-Step Solution

Answer

The given quadrilateral is a square.

Step 1. Given information

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

Step 2. Required to find

Show that the quadrilateral is a square.

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

Step 3. Finding the slopes of the sides of quadrilateral

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: $m_4=\frac{0-3}{0-1}$

$m_4=3$

Step 4. Showing adjacent sides are perpendicular

As $m_1\times m_2=-\frac{1}{3}\times 3=-1$

$m_2\times m_3=3\times -\frac{1}{3}=-1$

$m_3\times m_4=-\frac{1}{3}\times 3=-1$

$m_4\times m_1=3\times -\frac{1}{3}=-1$

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

Step 5. Showing lengths of all sides are equal

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:  $\sqrt{{(4-3)}^2+{(2-(-1\left)\right)}^2}=\sqrt{1+9}=\sqrt{10}$

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

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

All sides of the quadrilateral are equal.