Respuesta :
Given:
Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).
To find:
Whether the parallelogram ABCD is a rhombus, rectangle or square.
Solution:
Distance formula:
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, we get
[tex]AB=\sqrt{(-4-(-4))^2+(-1-7)^2}[/tex]
[tex]AB=\sqrt{(-4+4)^2+(-8)^2}[/tex]
[tex]AB=\sqrt{0+64}[/tex]
[tex]AB=8[/tex]
Similarly,
[tex]BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8[/tex]
[tex]CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8[/tex]
[tex]AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8[/tex]
All sides of parallelogram are equal.
[tex]AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}[/tex]
[tex]BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}[/tex]
Both diagonals are equal.
Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.
We know that, a square is special case of rectangles and rhombus.
So, parallelogram ABCD is a rhombus, rectangle or square. Therefore, the correct option is c.
