D.
The representation of this problem is shown below. To find the answer, we need to use the distance formula:
[tex]The \ \mathbf{distance} \ d \ between \ the \ \mathbf{points} \ (x_{1},y_{1}) \ and \ (x_{2},y_{2}) \ in \ the \ plane \ is:\\ \\ d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
So, for the first diagonal:
[tex]A(x_{1},y_{1})=A(-1,10) \\ \\ C(x_{2},y_{2})=C(1,8) \\ \\ \\ d_{1}=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \\ \\ d_{1}=\sqrt{[1-(-1)]^2+(8-10)^2}=2\sqrt{2}[/tex]
For the second diagonal:
[tex]B(x_{1},y_{1})=B(-4,5) \\ \\ D(x_{2},y_{2})=D(4,3) \\ \\ \\ d_{2}=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \\ \\ d_{2}=\sqrt{[4-(-4)]^2+(3-5)^2}=2\sqrt{17}[/tex]
So the diagonals aren't congruent. Are they perpendicular?
[tex]\vec{AC}=(1,8)-(-1,10)=(2,-2) \\ \\ \vec{BD}=(4,3)-(-4,5)=(8,-2)[/tex]
These two vectors will be perpendicular (hence the diagonals will be perpendicular) if and only if the dot product equals zero, so:
[tex](2,-2).(8,-2)=2(8)+(-2)(-2)=20\neq 0[/tex]
Thus, the diagonals aren't perpendicular. In conclusion: