Q51. - Algebra 2

Question

Determine whether quadrilateral $\mathbf{ABCD}$ with vertices $\mathbf{A}\left(-\mathbf{2},-\mathbf{1}\right)$, $\mathbf{B}\left(\mathbf{1},\mathbf{1}\right)$, $\mathbf{C}\left(\mathbf{3},-\mathbf{2}\right)$ and $\mathbf{D}\left(\mathbf{0},-\mathbf{4}\right)$ is a rectangle. Explain.

Step-by-Step Solution

Answer

The quadrilateral $ABCD$ is a rectangle.

Step 1 – State the concept

A quadrilateral whose internal angles are all $90°$ or equivalently, whose adjacent sides are all perpendicular is either a rectangle or a square, which is a special case of a rectangle. So, it is enough to check if the adjacent sides are all perpendicular to check if a quadrilateral is a rectangle.

 

The slope of a line passing through $\left(a,b\right)$ and $\left(c,d\right)$ is $\frac{d-b}{c-a}$.

 

The product of slopes of perpendicular lines is $-1$.

Step 2 – List the given data

The vertices of a quadrilateral $ABCD$ are $A\left(-2,-1\right)$, $B\left(1,1\right)$, $C\left(3,-2\right)$ and $D\left(0,-4\right)$.

Step 3 – Calculate the slopes

The slope of $AB$ is $\frac{1-\left(-1\right)}{1-\left(-2\right)}=\frac{2}{3}$

 

The slope of $BC$ is $\frac{-2-1}{3-1}=-\frac{3}{2}$

 

The slope of $CD$ is $\frac{-4-\left(-2\right)}{0-3}=\frac{2}{3}$

 

The slope of $AD$ is $\frac{-4-\left(-1\right)}{0-\left(-2\right)}=-\frac{3}{2}$

Step 4 – Check the perpendicularity of adjacent sides

The product of slopes of $AB$ and $BC$ is $\left(\frac{2}{3}\right)\left(-\frac{3}{2}\right)=-1$

 

The product of slopes of $BC$ and $CD$ is $\left(-\frac{3}{2}\right)\left(\frac{2}{3}\right)=-1$

 

The product of slopes of $CD$ and $AD$ is $\left(\frac{2}{3}\right)\left(-\frac{3}{2}\right)=-1$

 

The product of slopes of $AD$ and $AB$ is $\left(-\frac{3}{2}\right)\left(\frac{2}{3}\right)=-1$

 

This implies that the adjacent sides are all perpendicular to each other.

 

So, the quadrilateral $ABCD$ is a rectangle.