Among the statements:(S1) : If A(5, -1) and B | vedclass

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(D) Solution of statement - $1$:Let the third vertex be $C(h, k)$.The orthocentre $O$ is $(0, 0)$.Since $AO \perp BC$, the slope of $AO$ is $m_{AO} =

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Both are correct

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(D) Solution of statement - $1$:Let the third vertex be $C(h, k)$.The orthocentre $O$ is $(0, 0)$.Since $AO \perp BC$, the slope of $AO$ is $m_{AO} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.The slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.Since $m_{AO} \cdot m_{BC} = -1$, we have $(-\frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ ....$(1)$Since $BO \perp AC$, the slope of $BO$ is $m_{BO} = \frac{0 - 3}{0 - (-2)} = -\frac{3}{2}$.The slope of $AC$ is $m_{AC} = \frac{k - (-1)}{h - 5} = \frac{k + 1}{h - 5}$.Since $m_{BO} \cdot m_{AC} = -1$, we have $(-\frac{3}{2}) \cdot (\frac{k + 1}{h - 5}) = -1$ $\Rightarrow 3(k + 1) = 2(h - 5)$ $\Rightarrow 3k + 3 = 2h - 10$ $\Rightarrow 2h - 3k = 13$ ....$(2)$Solving equations $(1)$ and $(2)$, we get $h = -4$ and $k = -7$.So, the third vertex is $(-4, -7)$. Thus, statement $1$ is correct.Solution of statement - $2$:Given $2a, b, c$ are in $A.P.$, so $2b = 2a + c \Rightarrow 2a - 2b + c = 0$.The line equation is $ax + by + c = 0$.Comparing this with $2a - 2b + c = 0$, we can see that for $x = 2$ and $y = -2$, the equation $a(2) + b(-2) + c = 0$ holds true.Thus, the lines $ax + by + c = 0$ are concurrent at $(2, -2)$. Thus, statement $2$ is correct.

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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