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(A) The given equation of the pair of lines is $x^2-y^2-x+3y-2=0$. Factoring the expression,we get $(x-y+1)(x+y-2)=0$. Thus,the lines are $L_1: x-y+1=

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$\frac{5}{2}$

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(A) The given equation of the pair of lines is $x^2-y^2-x+3y-2=0$. Factoring the expression,we get $(x-y+1)(x+y-2)=0$. Thus,the lines are $L_1: x-y+1=0$ and $L_2: x+y-2=0$. Solving for the intersection point $P$: Adding the equations: $(x-y+1) + (x+y-2) = 2x-1=0 \implies x=\frac{1}{2}$. Substituting $x=\frac{1}{2}$ into $x+y-2=0$,we get $y=\frac{3}{2}$. So,$P = (\frac{1}{2}, \frac{3}{2})$. $\alpha$ is the square of the distance from the origin $(0,0)$ to $P$: $\alpha = (\frac{1}{2})^2 + (\frac{3}{2})^2 = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}$. $\beta$ is the product of the perpendicular distances from the origin to $L_1$ and $L_2$: $d_1 = \frac{|0-0+1|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$. $d_2 = \frac{|0+0-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. $\beta = d_1 	imes d_2 = \frac{1}{\sqrt{2}} 	imes \sqrt{2} = 1$. Therefore,$\alpha \beta = \frac{5}{2} 	imes 1 = \frac{5}{2}$.

If $\alpha$ represents the square of the distance between the origin and the point of intersection of the lines $x^2-y^2-x+3y-2=0$ and $\beta$ represents the product of the perpendicular distances from the origin to the pair of lines,then $\alpha \beta=$

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