Question

# The lines $$3x+y=0$$ may be perpendicular to one of the lines given by $$ax^2+2hxy+by^2=0$$.

Solution

## The line $$3x+y=0$$ may be perpendicular to one of the line given by $$a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=0$$Auxiliary equation$$=b{ m }^{ 2 }+2hm+a=0\longrightarrow (1)\\ 3x+y=0$$ slope$$({ m }_{ 1 })=-3$$ Hence slope  of the line perpendicular to given line.$$\therefore$$ Slope of one of the lines represented by the joint  equation is $$\cfrac { 1 }{ 3 } .$$ Now $$\cfrac { 1 }{ 3 }$$ must be one of the roots of equation$$b{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }+2h\left( \cfrac { 1 }{ 3 } \right) +a=0\\ b\cfrac { 1 }{ 9 } +\cfrac { 2h }{ 3 } +0$$$$b+6h+9a=0\quad$$(The required condition)Maths

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