Question

A quadrilateral has vertices at A(–5, 5), B(1, 8), C(4, 2), D(–2, –2). Use slope to determine if the quadrilateral is a rectangle. Show your work.

Answer 1

With the properties of rectangle in mind we must perform two verification. Verify to see if opposite sides are parallel and adjacent sides are perpendicular.

We need to determine the slope of each side, using the formula,

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

Slope  of AB

$m_{AB}=\frac{8-5}{1--5}$

$\Rightarrow m_{AB}=\frac{8-5}{1+5}$

$\Rightarrow m_{AB}=\frac{3}{6}$

$\Rightarrow m_{AB}=\frac{1}{2}$

Slope of BC

$m_{BC}=\frac{2-8}{4-1}$

$\Rightarrow m_{BC}=\frac{-6}{3}$

$\Rightarrow m_{BC}=-2$

Slope of CD

$m_{CD}=\frac{-2-2}{-2-4}$

$\Rightarrow m_{CD}=\frac{-2-2}{-2-4}$

$\Rightarrow m_{CD}=\frac{-4}{-6}$

$\Rightarrow m_{CD}=\frac{2}{3}$

Slope of AD

$m_{AD}=\frac{-2-5}{-2--5}$

$\Rightarrow m_{AD}=\frac{-2-5}{-2+5}$

$\Rightarrow m_{AD}=\frac{-7}{3}$

Verify Parallel sides

If the quadrilateral is a rectangle, then opposite sides should have the same slope. But

$m_{AD} = \frac{-7}{3} \neq m_{BC}=-2$

$m_{AB}=\frac{1}{2} \neq m_{CD}=\frac{2}{3}$

Verify Perpendicularity

And also the product of slopes of all sides with a common vertex should be -1. But

$m_{AB} \times m_{BC}=\frac{1}{2} \times -2=-1$

$m_{AB} \times m_{AD}=\frac{1}{2} \times -\frac{7}{3} \neq -1$

$\Rightarrow m_{CD} \times m_{AD}=\frac{2}{3} \times -\frac{7}{3} \neq -1$

$\Rightarrow m_{CD} \times m_{BC}=\frac{2}{3} \times -2 \neq -1$

Since the quadrilateral fails to satisfy all these conditions, the quadrilateral is not a rectangle