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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

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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.

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  • $\begingroup$ 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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