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`h(1+m^(2)) = m(a+b)``h(1-m^(2))=m(a-b)``h(1+m^(2))=m(a-b)`None of these

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BSolution :

Equation of pair of bisectors of angles between lines `ax^(2) +2hxy +by^(2) =0` is <br> `(x^(2)-y^(2))/(xy) =(a-b)/(h)` <br> `rArr h(x^(2)-y^(2)) =(a-b) xy` (i) <br> But `y = mx` is one of these lines, then it will satisfy it. Substituting `y = mx` in (i) <br> `h(x^(2)-m^(2)x^(2)) =(a-b) x.mx` <br> Dividing by `x^(2)`, we get `h (1-m^(2)) =m(a-b)`. **Explain Pair of straight lines with examples.**

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