Let P be the pair of lines represented by 2x^ | vedclass

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(D) The given equation of the pair of lines is $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$.Factoring the expression,we get $(2y - x - 3)(y - 2x) = 0$.Thus,the l

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$\gamma < \alpha < \beta$

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(D) The given equation of the pair of lines is $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$.Factoring the expression,we get $(2y - x - 3)(y - 2x) = 0$.Thus,the lines are $2y - x - 3 = 0$ and $y - 2x = 0$.Solving these equations simultaneously: $y = 2x$,so $2(2x) - x - 3 = 0 \implies 3x = 3 \implies x = 1$.Thus,the point of intersection is $(1, 2)$,so $\alpha = 1$.For statement (ii),the line passing through the origin is $y - 2x = 0$,which has a slope $m = 2$. Thus,$\beta = 2$.For statement (iii),the angular bisectors are given by $\frac{2y - x - 3}{\sqrt{2^2 + (-1)^2}} = \pm \frac{y - 2x}{\sqrt{1^2 + (-2)^2}}$.This simplifies to $2y - x - 3 = \pm(y - 2x)$.Case $1$: $2y - x - 3 = y - 2x \implies x + y - 3 = 0$.Case $2$: $2y - x - 3 = -y + 2x \implies 3x - 3y + 3 = 0 \implies x - y + 1 = 0$.The constant terms are $-3$ and $1$. Taking the standard form,$\gamma = -3$.Comparing the values: $\gamma = -3, \alpha = 1, \beta = 2$.Therefore,$\gamma < \alpha < \beta$.

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