The figure formed by the pairs of lines 6x^2+ | vedclass

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(C) The given equations are $6x^2+13xy+6y^2=0$ and $6x^2+13xy+6y^2+10x+10y+4=0$.Factorizing the first equation:$6x^2+9xy+4xy+6y^2=0$$3x(2x+3y)+2y(2x+3

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Rhombus

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(C) The given equations are $6x^2+13xy+6y^2=0$ and $6x^2+13xy+6y^2+10x+10y+4=0$.Factorizing the first equation:$6x^2+9xy+4xy+6y^2=0$$3x(2x+3y)+2y(2x+3y)=0$$(3x+2y)(2x+3y)=0$So,the lines are $L_1: 3x+2y=0$ and $L_2: 2x+3y=0$.Factorizing the second equation:$6x^2+13xy+6y^2+10x+10y+4=0$$(3x+2y+2)(2x+3y+2)=0$So,the lines are $L_3: 3x+2y+2=0$ and $L_4: 2x+3y+2=0$.The lines $L_1$ and $L_3$ are parallel,and $L_2$ and $L_4$ are parallel,so the figure is a parallelogram.The distance between parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.Distance between $L_1$ and $L_3$: $d_1 = \frac{|2-0|}{\sqrt{3^2+2^2}} = \frac{2}{\sqrt{13}}$.Distance between $L_2$ and $L_4$: $d_2 = \frac{|2-0|}{\sqrt{2^2+3^2}} = \frac{2}{\sqrt{13}}$.Since $d_1 = d_2$ and the lines are not perpendicular (slopes are $-3/2$ and $-2/3$),the figure is a rhombus.

The figure formed by the pairs of lines $6x^2+13xy+6y^2=0$ and $6x^2+13xy+6y^2+10x+10y+4=0$ is a

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