Question

Determine whether a figure with the given vertices is a rectangle using the Distance Formula.

Answer 1

Answer:

Distance of AB=Distance of CD

and Distance of BC=Distance of DA

Therefore the opposite sides of rectangle ABCD are congruent to each other

Step-by-step explanation:

Let ABCD be the rectangle

and the points A(4,-7), B(4,-2), C(0,-2), D(0,-7)

Now to verify that the given points can figure as a rectangle by using distance formula

Distance of AB

Let $(x_{1},y_{1}) ,(x_{2},y_{2})$ be the points A(4,-7), B(4,-2)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$d=\sqrt{(4-4)^2+(-2-(-7))^2}$

$d=\sqrt{(0)^2+(5)^2}$

$d=\sqrt{(5)^2}$

Therefore d=5units

Distance of BC

Let $(x_{1},y_{1}) ,(x_{2},y_{2})$ be the points  B(4,-2) and C(0,-2)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$d=\sqrt{(0-4)^2+(-2-(-2))^2}$

$d=\sqrt{(-4)^2+(0)^2}$

$d=\sqrt{16}$

Therefore d=4units

Distance of CD

Let $(x_{1},y_{1}) ,(x_{2},y_{2})$ be the points  C(0,-2) and D(0,-7)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$d=\sqrt{(0-0)^2+(-7-(-2))^2}$

$d=\sqrt{(0)^2+(-5)^2}$

$d=\sqrt{25}$

Therefore d=5units

Distance of DA

Let $(x_{1},y_{1}) ,(x_{2},y_{2})$ be the points  D(0,-7) and A(4,-7)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$d=\sqrt{(4-0)^2+(-7-(-7))^2}$

$d=\sqrt{(4)^2+(0)^2}$

$d=\sqrt{16}$

Therefore d=4units

Therefore Distance of AB=Distance of CD

and Distance of BC=Distance of DA

Therefore the opposite sides of rectangle ABCD are congruent to each other