The number of possible distinct straight line | vedclass

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(C) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through the point $(2,3)$,we have $\frac{2}{a} + \frac{3}{b

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(C) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through the point $(2,3)$,we have $\frac{2}{a} + \frac{3}{b} = 1$,which implies $2b + 3a = ab$. Given the area of the triangle is $12$ sq. units,we have $\frac{1}{2}|ab| = 12$,so $ab = \pm 24$. Case $I$: $ab = 24$. Substituting $b = \frac{24}{a}$ into $2b + 3a = ab$,we get $2(\frac{24}{a}) + 3a = 24$,which simplifies to $3a^2 - 24a + 48 = 0$,or $a^2 - 8a + 16 = 0$. This gives $(a-4)^2 = 0$,so $a = 4$ and $b = 6$. This provides $1$ line. Case $II$: $ab = -24$. Substituting $b = \frac{-24}{a}$ into $2b + 3a = ab$,we get $2(\frac{-24}{a}) + 3a = -24$,which simplifies to $3a^2 + 24a - 48 = 0$,or $a^2 + 8a - 16 = 0$. The roots are $a = \frac{-8 \pm \sqrt{64 - 4(1)(-16)}}{2} = -4 \pm 4\sqrt{2}$. Since $b = \frac{-24}{a}$,each value of $a$ gives a distinct value of $b$. This provides $2$ additional lines. Thus,the total number of possible lines is $1 + 2 = 3$.

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	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$

The number of possible distinct straight lines passing through $(2,3)$ and forming a triangle with the coordinate axes whose area is $12$ sq. units is:

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