If two diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$, then parallelogram represents
(a) rectangle
(b) rhombus
(c) square
(d) cyclic quadrilateral

$\bf{Attempts}$ with the help of slope

Slope of $x+3y=4$ is $\displaystyle m_{1} = -\frac{1}{3}$ and slope of $6x-2y=7$ is $\displaystyle m_{2} = -\frac{6}{-2} = 3$

So we have $m_{1}\cdot m_{2} = -1$ . So parallelogram represent Rhombus and Square.

But answer given is only Rhombus, could someone explain me the reason, Thanks


The two diagonals of the parallelogram being perpendicular to each other only defines a rhombus, not a square in particular. A square is a special case of a rhombus, with equal angles.

  • $\begingroup$ a square is also a parallelogram, but a Rhombus must not be a square $\endgroup$ – Dr. Sonnhard Graubner Jul 18 '17 at 15:39
  • $\begingroup$ @Dr.SonnhardGraubner Trite, whaddya know? A rhombus may be a square. $\endgroup$ – Parcly Taxel Jul 18 '17 at 15:44
  • $\begingroup$ Thanks Parcly Taxel, can we say that option $(c)$ is also true. $\endgroup$ – DXT Jul 18 '17 at 15:45
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    $\begingroup$ @DurgeshTiwari By that I mean you can draw non-square rhombuses with the given diagonals, so (c) is not true in all cases. $\endgroup$ – Parcly Taxel Jul 18 '17 at 15:50
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    $\begingroup$ @DurgeshTiwari In other words, there’s not enough information given to say for sure that it’s a square. $\endgroup$ – amd Jul 18 '17 at 19:12

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