Exercise 27 Page 794

Let's plot the given points on a coordinate plane and graph the quadrilateral.

We can see that the quadrilateral looks like a rhombus or a square. Recall that a rhombus is a parallelogram with all four sides congruent. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. Let's check if this is the case by finding the slopes using the Slope Formula.

Side	Points	$\frac{y_2-y_1}{x_2-x_1}$	Simplify
$\overline{XW}$	$X(0,3),$ $W(3,0)$	$\frac{0-3}{3-0}$	$\text{-}1$
$\overline{WZ}$	$W(3,0),$ $Z(0,\text{-}3)$	$\frac{\text{-}3-0}{0-3}$	$1$
$\overline{ZY}$	$Z(0,\text{-}3),$ $Y(\text{-}3,0)$	$\frac{0-(\text{-}3)}{\text{-}3-0}$	$\text{-}1$
$\overline{YX}$	$Y(\text{-}3,0),$ $X(0,3)$	$\frac{3-0}{0-(\text{-}3)}$	$1$

We can tell that the slopes of the opposite sides of our quadrilateral are equal. Therefore, both pairs of opposite sides are parallel and the quadrilateral is a parallelogram. We can also tell that the consecutive sides are perpendicular, as their slopes are opposite reciprocals. Therefore, our quadrilateral is either a rectangle or a square. To check, we can find the lengths of its sides using the Distance Formula.

Side	Points	$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$	Simplify
$\overline{XW}$	$X(0,3),$ $W(3,0)$	$\sqrt{(3-0{)}^2+(0-3{)}^2}$	$\sqrt{18}$
$\overline{WZ}$	$W(3,0),$ $Z(0,\text{-}3)$	$\sqrt{(0-3{)}^2+(\text{-}3-0{)}^2}$	$\sqrt{18}$
$\overline{ZY}$	$Z(0,\text{-}3),$ $Y(\text{-}3,0)$	$\sqrt{(\text{-}3-0{)}^2+(0-(\text{-}3){)}^2}$	$\sqrt{18}$
$\overline{YX}$	$Y(\text{-}3,0),$ $X(0,3)$	$\sqrt{(0-(\text{-}3){)}^2+(3-0{)}^2}$	$\sqrt{18}$

Our parallelogram has four congruent sides and these sides are perpendicular to each other. Therefore, it is a square. We want to find its area. The area of a square equals its side length squared, $A=s^2.$ In our case $s=\sqrt{18}.$ The are of the given quadrilateral is $18$ square units.