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(C) The pair of lines $x^2+xy-2y^2=0$ can be factored as $(x+2y)(x-y)=0$. The lines are $L_1: x+2y=0$ and $L_2: x-y=0$.The pair of lines $3y^2-5xy-2x^

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(C) The pair of lines $x^2+xy-2y^2=0$ can be factored as $(x+2y)(x-y)=0$. The lines are $L_1: x+2y=0$ and $L_2: x-y=0$.The pair of lines $3y^2-5xy-2x^2=0$ can be rewritten as $2x^2+5xy-3y^2=0$,which factors as $(2x-y)(x+3y)=0$. The lines are $L_3: 2x-y=0$ and $L_4: x+3y=0$.We are given that one line from the first pair is perpendicular to one line from the second pair.The slopes of the lines are $m_1 = -1/2$,$m_2 = 1$,$m_3 = 2$,and $m_4 = -1/3$.Since $m_2 	imes m_4 = 1 	imes (-1/3) 
eq -1$ and $m_1 	imes m_3 = (-1/2) 	imes 2 = -1$,the lines $L_1$ and $L_3$ are perpendicular.The remaining two lines are $L_2: x-y=0$ and $L_4: x+3y=0$.The combined equation of these two lines is $(x-y)(x+3y) = x^2+3xy-xy-3y^2 = x^2+2xy-3y^2=0$.Comparing this with $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=2$,and $b=-3$.Thus,$a+2h+b = 1+2-3 = 0$.

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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