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I started to solve the question by taking the sides of rectangle ABCD then added a midpoint in the rectangle and divided the rectangle in diagonal then found out the midpoint of diagonals AC and BD (Which was the most possible thing i could do in the question).But at the page where all the answers are written rather than (4,2) since both diagonals bisect each other is written [Hence by converse of Pythagoras' theorem

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The vector from $(0,-1)$ to $(4,-3)$ is $(4,-2)$, as is the vector from $(4,7)$ to $(8,5)$. You can check the other two sides are $(4,8)$ This establishes we have a parallelogram. Then since $(4,-2)\cdot (4,8)=0$ they are perpendicular-a rectangle.

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hint:let A=(0,-1) B=(4,-3 )C=(8,5) D= (4,7) be angular point of rectangle prove |AB|=|Cd| |AC|=|BD| and these vector are parallel {AB parallel with CD and AC parallel with BD} and show that inner product of AB ,BD is zero its mean AB$\bot$BD (=0 similarly indicate that $<AC,CD>=0 <AB,BC>=0 ,<BD,DC>=0$

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