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=[tex]\sqrt{(2+4)^{2}+(-5-7)^{2}}[/tex]
=[tex]\sqrt{36+144}[/tex]
=[tex]3\sqrt{20}[/tex]
Also, QS= [tex]\sqrt{(5+7)^{2}+(4+2)^{2}}[/tex]
=[tex]\sqrt{144+36}[/tex]
=[tex]3\sqrt{20}[/tex]
Therefore, PR is congruent to QS that is PR≅QS.
Slope of PR= [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
=[tex]\frac{-5-7}{2+4}=\frac{-12}{6}=-2[/tex]
Slope of QS=[tex]\frac{-2-4}{-7-5}=\frac{-6}{-12}=\frac{1}{2}[/tex]
Thus, PR⊥QS.
Now, Mid point of PR=[tex](\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})[/tex]
=[tex](\frac{-4+2}{2}, \frac{7-5}{2})[/tex]
=[tex](-1,1)[/tex]
Also, mid point of QS=[tex](\frac{5-7}{2}, \frac{-2+4}{2})[/tex]
=[tex](-1,1)[/tex]
Therefore, (-1,1) is the mid point of both PR and QS, so PR and QS bisect each other.
Hence, option (A) is correct.
