The pairs of straight lines x^2-3xy+2y^2=0 an | vedclass

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(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$. Factorizing the first equation: $(x-2y)(x-y)=0$,which represents lines $x-

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parallelogram

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(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$. Factorizing the first equation: $(x-2y)(x-y)=0$,which represents lines $x-2y=0$ and $x-y=0$. Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which represents lines $x-2y+2=0$ and $x-y-1=0$. Since the lines $x-2y=0$ and $x-2y+2=0$ are parallel,and the lines $x-y=0$ and $x-y-1=0$ are parallel,the figure formed is a parallelogram. To check if it is a rectangle,we find the angle between $x-2y=0$ and $x-y=0$. The slopes are $m_1 = 1/2$ and $m_2 = 1$. Since $m_1 	imes m_2 
eq -1$,the angle is not $90^{\circ}$. Thus,it is a parallelogram.

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

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