Question

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

A(4, –7), B(4, –2), C(0, –2), D(0, –7)

Answer 1

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

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=√((4-4)^2+(-2-(-7))^2)`

`d=√((0)^2+(5)^2)`

`d=√((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=√((0-4)^2+(-2-(-2))^2)`

`d=√((-4)^2+(0)^2)`

`d=√(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=√((0-0)^2+(-7-(-2))^2)`

`d=√((0)^2+(-5)^2)`

`d=√(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=√((4-0)^2+(-7-(-7))^2)`

`d=√((4)^2+(0)^2)`

`d=√(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