The line x - 2y = 0 will be a bisector of the | vedclass

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(C) The equation of the pair of lines is ${x^2} - 2hxy - 2{y^2} = 0$. Comparing this with $ax^2 + 2h'xy + by^2 = 0$,we have $a = 1$,$2h' = -2h$,and $b

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$-2$

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(C) The equation of the pair of lines is ${x^2} - 2hxy - 2{y^2} = 0$. Comparing this with $ax^2 + 2h'xy + by^2 = 0$,we have $a = 1$,$2h' = -2h$,and $b = -2$. The equation of the bisectors of the angles between the lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h'}$. Substituting the values,we get $\frac{x^2 - y^2}{1 - (-2)} = \frac{xy}{-h} \Rightarrow \frac{x^2 - y^2}{3} = \frac{xy}{-h}$. This simplifies to $-h(x^2 - y^2) = 3xy$,or $hx^2 + 3xy - hy^2 = 0$. Since $x - 2y = 0$ is a bisector,it must satisfy this equation. Substituting $x = 2y$,we get $h(2y)^2 + 3(2y)y - hy^2 = 0$. $4hy^2 + 6y^2 - hy^2 = 0 \Rightarrow 3hy^2 + 6y^2 = 0$. $3y^2(h + 2) = 0$. For this to hold for all $y$,$h + 2 = 0$,so $h = -2$.

The line $x - 2y = 0$ will be a bisector of the angle between the lines represented by the equation ${x^2} - 2hxy - 2{y^2} = 0$,if $h = $

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