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:
![D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=D%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Using distance formula, we get
![AB=\sqrt{(-4-(-4))^2+(-1-7)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-4-%28-4%29%29%5E2%2B%28-1-7%29%5E2%7D)
![AB=\sqrt{(-4+4)^2+(-8)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-4%2B4%29%5E2%2B%28-8%29%5E2%7D)
![AB=\sqrt{0+64}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B0%2B64%7D)
![AB=8](https://tex.z-dn.net/?f=AB%3D8)
Similarly,
![BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%28-1-%28-1%29%29%5E2%2B%2812-%28-4%29%29%5E2%7D%3D8)
![CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8](https://tex.z-dn.net/?f=CD%3D%5Csqrt%7B%287-%28-1%29%29%5E2%2B%28-12-%28-12%29%29%5E2%7D%3D8)
![AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8](https://tex.z-dn.net/?f=AD%3D%5Csqrt%7B%287-7%29%5E2%2B%28-12-%28-4%29%29%5E2%7D%3D8)
All sides of parallelogram are equal.
![AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B%28-1-7%29%5E2%2B%28-12-%28-4%29%29%5E2%7D%3D8%5Csqrt%7B2%7D)
![BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}](https://tex.z-dn.net/?f=BD%3D%5Csqrt%7B%287-%28-1%29%29%5E2%2B%28-12-%28-4%29%29%5E2%7D%3D8%5Csqrt%7B2%7D)
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.