One line of the pair of lines x^2+xy-2y^2=0 i | vedclass

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(C) The given pairs of lines are $x^2+xy-2y^2=0$ and $2x^2+5xy-3y^2=0$. Factoring the first pair: $(x+2y)(x-y)=0$. The lines are $L_1: x+2y=0$ and $L_

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(C) The given pairs of lines are $x^2+xy-2y^2=0$ and $2x^2+5xy-3y^2=0$. Factoring the first pair: $(x+2y)(x-y)=0$. The lines are $L_1: x+2y=0$ and $L_2: x-y=0$. Factoring the second pair: $(2x-y)(x+3y)=0$. The lines are $L_3: 2x-y=0$ and $L_4: x+3y=0$. We check for perpendicularity: The slope of $L_1$ is $m_1 = -1/2$. The slope of $L_3$ is $m_3 = 2$. Since $m_1 	imes m_3 = -1$,$L_1$ and $L_3$ are perpendicular. The remaining lines are $L_2: x-y=0$ and $L_4: x+3y=0$. The combined equation of these two lines is $(x-y)(x+3y)=0$. Expanding this: $x^2+3xy-xy-3y^2 = x^2+2xy-3y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=2$,and $b=-3$. Therefore,$a+2h+b = 1+2-3 = 0$.

One line of the pair of lines $x^2+xy-2y^2=0$ is perpendicular to one line of the pair of lines $3y^2-5xy-2x^2=0$. If the combined equation of the two lines other than those two perpendicular lines is $ax^2+2hxy+by^2=0$,then $a+2h+b=$

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