Respuesta :
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 [tex](x_{1},y_{1}) ,(x_{2},y_{2})[/tex] be the points A(4,-7), B(4,-2)
[tex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
[tex]d=\sqrt{(4-4)^2+(-2-(-7))^2}[/tex]
[tex]d=\sqrt{(0)^2+(5)^2}[/tex]
[tex]d=\sqrt{(5)^2}[/tex]
Therefore d=5units
Distance of BC
Let [tex](x_{1},y_{1}) ,(x_{2},y_{2})[/tex] be the points B(4,-2) and C(0,-2)
[tex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
[tex]d=\sqrt{(0-4)^2+(-2-(-2))^2}[/tex]
[tex]d=\sqrt{(-4)^2+(0)^2}[/tex]
[tex]d=\sqrt{16}[/tex]
Therefore d=4units
Distance of CD
Let [tex](x_{1},y_{1}) ,(x_{2},y_{2})[/tex] be the points C(0,-2) and D(0,-7)
[tex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
[tex]d=\sqrt{(0-0)^2+(-7-(-2))^2}[/tex]
[tex]d=\sqrt{(0)^2+(-5)^2}[/tex]
[tex]d=\sqrt{25}[/tex]
Therefore d=5units
Distance of DA
Let [tex](x_{1},y_{1}) ,(x_{2},y_{2})[/tex] be the points D(0,-7) and A(4,-7)
[tex]d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
[tex]d=\sqrt{(4-0)^2+(-7-(-7))^2}[/tex]
[tex]d=\sqrt{(4)^2+(0)^2}[/tex]
[tex]d=\sqrt{16}[/tex]
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
