If r(1 - m^2) + m(p - q) = 0,then a bisector | vedclass

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(C) The equation of the bisectors of the angle between the lines represented by $px^2 - 2rxy + qy^2 = 0$ is given by:$\frac{x^2 - y^2}{p - q} = \frac{

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$y = mx$

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(C) The equation of the bisectors of the angle between the lines represented by $px^2 - 2rxy + qy^2 = 0$ is given by:$\frac{x^2 - y^2}{p - q} = \frac{xy}{-r}$ .....$(i)$To check if $y = mx$ is a bisector,we substitute $y = mx$ into equation $(i)$:$\frac{x^2 - (mx)^2}{p - q} = \frac{x(mx)}{-r}$$\frac{x^2(1 - m^2)}{p - q} = \frac{mx^2}{-r}$Dividing by $x^2$ (assuming $x 
eq 0$):$\frac{1 - m^2}{p - q} = \frac{m}{-r}$$-r(1 - m^2) = m(p - q)$$r(1 - m^2) + m(p - q) = 0$Since this matches the given condition,$y = mx$ is a bisector.

If $r(1 - m^2) + m(p - q) = 0$,then a bisector of the angle between the lines represented by the equation $px^2 - 2rxy + qy^2 = 0$ is

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