Question

quadrilateral abcd has the following vertices: • a(0,6) • b(3,5) • c(0, - 4) • d(-3,-3) also, ∠a is a right angle. is quadrilateral abcd a rectangle, and why? choose 1 answer:

Explanation:

Step1: Calculate the slopes of adjacent sides

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Slope of $AB$ with $A(0,6)$ and $B(3,5)$: $m_{AB}=\frac{5 - 6}{3-0}=-\frac{1}{3}$.
Slope of $AD$ with $A(0,6)$ and $D(-3,-3)$: $m_{AD}=\frac{-3 - 6}{-3-0}=3$.
Since $m_{AB}\times m_{AD}=-\frac{1}{3}\times3=- 1$, $\angle A$ is a right - angle.
Slope of $BC$ with $B(3,5)$ and $C(0,-4)$: $m_{BC}=\frac{-4 - 5}{0 - 3}=3$.
Slope of $CD$ with $C(0,-4)$ and $D(-3,-3)$: $m_{CD}=\frac{-3+4}{-3 - 0}=-\frac{1}{3}$.

Step2: Check for parallel and perpendicular sides

Two lines are parallel if their slopes are equal and perpendicular if the product of their slopes is - 1.
$m_{AB}=m_{CD}=-\frac{1}{3}$, so $AB\parallel CD$.
$m_{AD}=m_{BC}=3$, so $AD\parallel BC$.
Also, $m_{AB}\times m_{BC}=-\frac{1}{3}\times3=-1$, $m_{BC}\times m_{CD}=3\times(-\frac{1}{3})=-1$, $m_{CD}\times m_{AD}=-\frac{1}{3}\times3=-1$.
Since opposite sides are parallel and adjacent sides are perpendicular, quadrilateral $ABCD$ is a rectangle.

Answer:

Yes, because opposite sides are parallel and adjacent sides are perpendicular.