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(D) The given line is $x+y-2=0$. To find the $X$-intercept '$a$',set $y=0$: $x+0-2=0 \Rightarrow x=2$. Thus,$a=2$. The line can be written in intercep

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$4$

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(D) The given line is $x+y-2=0$. To find the $X$-intercept '$a$',set $y=0$: $x+0-2=0 \Rightarrow x=2$. Thus,$a=2$. The line can be written in intercept form as $\frac{x}{2} + \frac{y}{2} = 1$. The normal form of a line is $x \cos \beta + y \sin \beta = p$. Comparing $x+y=2$ with $x \cos \beta + y \sin \beta = p$,we divide by $\sqrt{1^2+1^2} = \sqrt{2}$: $\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = \frac{2}{\sqrt{2}} = \sqrt{2}$. Here,$\cos \beta = \frac{1}{\sqrt{2}}$ and $\sin \beta = \frac{1}{\sqrt{2}}$,so $\beta = 45^{\circ}$. The perpendicular distance $p = \sqrt{2}$. Now,calculate $a 	an \beta + p^2$: $a 	an \beta + p^2 = 2 	an(45^{\circ}) + (\sqrt{2})^2 = 2(1) + 2 = 4$.

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