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

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(C) The given equations 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 gives lines $L_1: x-2y=0$

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parallelogram

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(C) The given equations 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 gives lines $L_1: x-2y=0$ and $L_2: x-y=0$. Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which gives lines $L_3: x-2y+2=0$ and $L_4: x-y-1=0$. Since $L_1 \parallel L_3$ and $L_2 \parallel L_4$,the figure formed is a parallelogram. The slopes of the lines are $m_1 = 1/2$ and $m_2 = 1$. Since $m_1 	imes m_2 
eq -1$,the angle between the lines is not $90^{\circ}$. Thus,the figure 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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