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Question

The diagonals of a parallelogram $$PQRS$$ are along the lines $$x + 3y = 4$$ and $$6x -2y = 7$$, then $$PQRS$$  must be a


A
rectangle
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B
square
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C
cyclic quadrilateral
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D
rhombus
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Solution

The correct option is C rhombus
Diagonals of a parallelogram $$PQRS$$ are along the lines $$x + 3y = 4$$ and $$6x -2y = 7.$$

For the line $$x+3y=4$$. Hence, $$m_1 = -\cfrac 13$$.
For the line $$6x+-2y=7$$. Hence, $$m_2 = 3$$ 

$$\therefore m_1 \times m_2=-1.$$

Therefore, the diagonals of the parallelogram are perpendicular to each 
other. 

The given parallelogram is a rhombus. 

Mathematics

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