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(C) . Let $Y$-intercept be $a$,then $X$-intercept is $2a$. Equation: $\frac{x}{2a} + \frac{y}{a} = 1 \Rightarrow x+2y=2a$. Since it passes through $(4

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$A-(IV), B-(II), C-(V), D-(I)$

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(C) . Let $Y$-intercept be $a$,then $X$-intercept is $2a$. Equation: $\frac{x}{2a} + \frac{y}{a} = 1 \Rightarrow x+2y=2a$. Since it passes through $(4,3)$,$4+2(3)=2a \Rightarrow 2a=10$. Equation: $x+2y-10=0$ (Option $IV$).$B$. Centroid $G = (\frac{1+3+6}{3}, \frac{1+3-6}{3}) = (\frac{10}{3}, -\frac{2}{3})$. Circumcentre $O$: Midpoint of $AB$ is $(2,2)$,slope $m_{AB} = 1$,perp slope $-1$. Line $x+y=4$. Midpoint of $BC$ is $(4.5, -1.5)$,slope $m_{BC} = \frac{-6-3}{6-3} = -3$,perp slope $1/3$. Line $y+1.5 = \frac{1}{3}(x-4.5) \Rightarrow x-3y=9$. Solving $x+y=4$ and $x-3y=9$ gives $O = (\frac{21}{4}, -\frac{5}{4})$. Line through $G$ and $O$ is $7x+23y-8=0$ (Option $II$).$C$. Line perpendicular to $x-y+2=0$ has slope $-1$. Equation: $y-0 = -1(x - (-3/5)) \Rightarrow y = -x - 3/5 \Rightarrow 5x+5y+3=0$ (Option $V$).$D$. Normal form: $x \cos 45^{\circ} + y \sin 45^{\circ} = 2 \Rightarrow x(\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) = 2 \Rightarrow x+y-2\sqrt{2}=0$ (Option $I$).

List-$I$	List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept	$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$	$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$	$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$

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