Answer:
Option (A)
Step-by-step explanation:
The vertices of square PQRS are P(−4, 7), Q(5, 4), R(2,−5) and S(−7,−2).
Now, Join the diagonals PR and QS,
now, PR=![\sqrt{(2+4)^{2}+(-5-7)^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%282%2B4%29%5E%7B2%7D%2B%28-5-7%29%5E%7B2%7D%7D)
=![\sqrt{36+144}](https://tex.z-dn.net/?f=%5Csqrt%7B36%2B144%7D)
=![3\sqrt{20}](https://tex.z-dn.net/?f=3%5Csqrt%7B20%7D)
Also, QS= ![\sqrt{(5+7)^{2}+(4+2)^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%285%2B7%29%5E%7B2%7D%2B%284%2B2%29%5E%7B2%7D%7D)
=![\sqrt{144+36}](https://tex.z-dn.net/?f=%5Csqrt%7B144%2B36%7D)
=![3\sqrt{20}](https://tex.z-dn.net/?f=3%5Csqrt%7B20%7D)
Therefore, PR is congruent to QS that is PR≅QS.
Slope of PR= ![\frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
=![\frac{-5-7}{2+4}=\frac{-12}{6}=-2](https://tex.z-dn.net/?f=%5Cfrac%7B-5-7%7D%7B2%2B4%7D%3D%5Cfrac%7B-12%7D%7B6%7D%3D-2)
Slope of QS=![\frac{-2-4}{-7-5}=\frac{-6}{-12}=\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-2-4%7D%7B-7-5%7D%3D%5Cfrac%7B-6%7D%7B-12%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Thus, PR⊥QS.
Now, Mid point of PR=![(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})](https://tex.z-dn.net/?f=%28%5Cfrac%7Bx_%7B1%7D%2Bx_%7B2%7D%7D%7B2%7D%2C%20%5Cfrac%7By_%7B1%7D%2By_%7B2%7D%7D%7B2%7D%29)
=![(\frac{-4+2}{2}, \frac{7-5}{2})](https://tex.z-dn.net/?f=%28%5Cfrac%7B-4%2B2%7D%7B2%7D%2C%20%5Cfrac%7B7-5%7D%7B2%7D%29)
=![(-1,1)](https://tex.z-dn.net/?f=%28-1%2C1%29)
Also, mid point of QS=![(\frac{5-7}{2}, \frac{-2+4}{2})](https://tex.z-dn.net/?f=%28%5Cfrac%7B5-7%7D%7B2%7D%2C%20%5Cfrac%7B-2%2B4%7D%7B2%7D%29)
=![(-1,1)](https://tex.z-dn.net/?f=%28-1%2C1%29)
Therefore, (-1,1) is the mid point of both PR and QS, so PR and QS bisect each other.
Hence, option (A) is correct.