The diagonal of a parallelogram PQRS are along the line $x+3y=4$ and $6x-2y=7$ . Then which geometrical figure (square, rhombus, rectangle, a cyclic quadrilateral) is PQRS?

I know the slopes are perpendicular, so the diagonals are perpendicular.

I got the same question in my exam and it was under the section of multiple options correct questions. Ideally, the answer should be square, rhombus and cyclic quadrilateral. But, we can argue by saying that the condition for rhombus will always satisfy for any parallelogram while condition for square and cyclic quadrilateral will satisfy only for specific parallelogram.

So the answer should be marked as square, parallelogram and rhombus or only rhombus.

  • $\begingroup$ The parallelogram must be a rhombus, it can be a specific rhombus, namely a square, in which case it is also a rectangle and a cyclic quadrilateral. So depending on the problem poser's intention, either only rhombus is correct or all answers are correct ... $\endgroup$ – Hagen von Eitzen Jul 16 '19 at 13:12
  • $\begingroup$ I am sorry. There wasn't a should in it. I have edited the question accordingly. $\endgroup$ – Aditya Jain Jul 16 '19 at 13:14
  • $\begingroup$ @HagenvonEitzen I don't think rectangle can ever be correct as its diagonals aren't perpendicular. $\endgroup$ – Aditya Jain Jul 16 '19 at 13:20
  • $\begingroup$ @AdityaJain They are if it's a square. $\endgroup$ – eyeballfrog Jul 16 '19 at 15:01

A special case of parallelogram (sides unequal)is the rhombus (when sides are equal). A special case of rhombus is a square (when diagonals are also equal).

A special case of parallelogram (oblique) is the rectangle (when angles are all equal to $90^{\circ}.)$ A special case of rectangle is a square. (when sides are equal).

The square is a special case of cyclic quadrilaterals when sides are all equal. Then each angle becomes equal to $90^{\circ}$.


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