If y = mx is one of the bisectors of the angl | vedclass

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(C) The equation of the pair of lines is $ax^2 - 2hxy + by^2 = 0$. The equation of the bisectors of the angles between these lines is given by $\frac{

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$h(1 - m^2) + m(a - b) = 0$

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(C) The equation of the pair of lines is $ax^2 - 2hxy + by^2 = 0$. The equation of the bisectors of the angles between these lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{-h}$.This can be rewritten as $h(x^2 - y^2) + (a - b)xy = 0$.Since $y = mx$ is one of the bisectors,it must satisfy the equation of the bisectors.Substituting $y = mx$ into the equation $h(x^2 - y^2) + (a - b)xy = 0$,we get:$h(x^2 - (mx)^2) + (a - b)x(mx) = 0$$h(x^2 - m^2x^2) + (a - b)mx^2 = 0$Dividing by $x^2$ (assuming $x 
eq 0$):$h(1 - m^2) + m(a - b) = 0$.

If $y = mx$ is one of the bisectors of the angle between the lines $ax^2 - 2hxy + by^2 = 0$,then

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