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=√((x_2-x_1)^2+(y_2-y_1)^2)](https://img.qammunity.org/2021/formulas/mathematics/college/c00i76k277fqbp0j3yzz4iqutiurlwhml8.png)
Using distance formula, we get
![AB=√((-4-(-4))^2+(-1-7)^2)](https://img.qammunity.org/2021/formulas/mathematics/college/voti2fzarjp3ck6zd90u3ngmovpv8q7wfa.png)
![AB=√((-4+4)^2+(-8)^2)](https://img.qammunity.org/2021/formulas/mathematics/college/b3m6s1hd5hak67xij8stnv2uwbyi3vn7pu.png)
![AB=√(0+64)](https://img.qammunity.org/2021/formulas/mathematics/college/6t04opkaeaa2of182o54zbx1iwj3zzyl1v.png)
![AB=8](https://img.qammunity.org/2021/formulas/mathematics/college/hprym8fnh03ccavcr8mb9h7g3so2mj5ul4.png)
Similarly,
![BC=√((-1-(-1))^2+(12-(-4))^2)=8](https://img.qammunity.org/2021/formulas/mathematics/college/tzlgg0gkf82iqe48vqvk7kg0kyr7utzi55.png)
![CD=√((7-(-1))^2+(-12-(-12))^2)=8](https://img.qammunity.org/2021/formulas/mathematics/college/t2asj6nu5aezv5nsxwhq6750t0vi09u1lr.png)
![AD=√((7-7)^2+(-12-(-4))^2)=8](https://img.qammunity.org/2021/formulas/mathematics/college/o2sq7y29yxzz24dvzmbqbdqu8b7jcr6462.png)
All sides of parallelogram are equal.
![AC=√((-1-7)^2+(-12-(-4))^2)=8√(2)](https://img.qammunity.org/2021/formulas/mathematics/college/1i37p0snlv7o8od2upqlv760qs2tnczwna.png)
![BD=√((7-(-1))^2+(-12-(-4))^2)=8√(2)](https://img.qammunity.org/2021/formulas/mathematics/college/xzgko02t9yb85xbdj55te149g29lfvxa2r.png)
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.