For all alpha, beta in R and alpha beta 0,t | vedclass

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(C) Given the line equation: $\alpha x + \beta y + \sqrt{\alpha \beta} = 0$. Dividing the equation by $-\sqrt{\alpha \beta}$,we get: $\frac{\alpha x}{

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forms a triangle of constant area with coordinate axes

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(C) Given the line equation: $\alpha x + \beta y + \sqrt{\alpha \beta} = 0$. Dividing the equation by $-\sqrt{\alpha \beta}$,we get: $\frac{\alpha x}{-\sqrt{\alpha \beta}} + \frac{\beta y}{-\sqrt{\alpha \beta}} = 1$ $\Rightarrow \frac{\sqrt{\alpha}}{\sqrt{\beta}} x + \frac{\sqrt{\beta}}{\sqrt{\alpha}} y = -1$. The $x$-intercept is $a = -\sqrt{\frac{\beta}{\alpha}}$ and the $y$-intercept is $b = -\sqrt{\frac{\alpha}{\beta}}$. The area of the triangle formed by the line with the coordinate axes is given by $Area = \frac{1}{2} |ab|$. $Area = \frac{1}{2} |(-\sqrt{\frac{\beta}{\alpha}}) 	imes (-\sqrt{\frac{\alpha}{\beta}})| = \frac{1}{2} |1| = \frac{1}{2}$ square units. Since the area is $\frac{1}{2}$,which is independent of $\alpha$ and $\beta$,the line forms a triangle of constant area with the coordinate axes.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

For all $\alpha, \beta \in R$ and $\alpha \beta > 0$,the line $\alpha x + \beta y + \sqrt{\alpha \beta} = 0$ is such that it

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