If the lines x + 3y = 4 and 6x - 2y = 7 are t | vedclass

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(D) Let the equations of the diagonals be $L_1: x + 3y = 4$ and $L_2: 6x - 2y = 7$. First,we find the slopes of these lines. For $L_1$,$3y = -x + 4 \i

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Rhombus

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(D) Let the equations of the diagonals be $L_1: x + 3y = 4$ and $L_2: 6x - 2y = 7$. First,we find the slopes of these lines. For $L_1$,$3y = -x + 4 \implies y = -\frac{1}{3}x + \frac{4}{3}$,so slope $m_1 = -\frac{1}{3}$. For $L_2$,$2y = 6x - 7 \implies y = 3x - \frac{7}{2}$,so slope $m_2 = 3$. Since $m_1 	imes m_2 = (-\frac{1}{3}) 	imes 3 = -1$,the diagonals are perpendicular to each other. $A$ parallelogram whose diagonals are perpendicular is a rhombus. Therefore,$PQRS$ is a rhombus.

If the lines $x + 3y = 4$ and $6x - 2y = 7$ are the diagonals of a parallelogram $PQRS$,then $PQRS$ is a:

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