The number of possible straight lines passing | vedclass

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(C) Let the equation of the line forming a triangle with the coordinate axes be $\frac{x}{a} + \frac{y}{b} = 1$.The area of the triangle is $\frac{1}{

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$3$

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(C) Let the equation of the line forming a triangle with the coordinate axes be $\frac{x}{a} + \frac{y}{b} = 1$.The area of the triangle is $\frac{1}{2} |ab| = 12$,so $|ab| = 24$.Since the line passes through $(2,3)$,we have $\frac{2}{a} + \frac{3}{b} = 1$,which implies $2b + 3a = ab$.Case $1$: $ab = 24$. Then $3a + 2b = 24$. Substituting $b = \frac{24-3a}{2}$ into $ab=24$ gives $a(24-3a) = 48$,or $3a^2 - 24a + 48 = 0$,so $a^2 - 8a + 16 = 0$. This gives $a=4, b=6$ ($1$ solution).Case $2$: $ab = -24$. Then $3a + 2b = -24$ (if the line is in the second or fourth quadrant).Substituting $b = \frac{-24-3a}{2}$ into $ab = -24$ gives $a(-24-3a) = -48$,or $3a^2 + 24a - 48 = 0$,so $a^2 + 8a - 16 = 0$.The discriminant is $D = 64 - 4(1)(-16) = 128 > 0$,yielding $2$ distinct real values for $a$,and thus $2$ distinct lines.Case $3$: $ab = -24$ and $3a + 2b = 24$. Substituting $b = \frac{24-3a}{2}$ into $ab = -24$ gives $a(24-3a) = -48$,or $3a^2 - 24a - 48 = 0$,so $a^2 - 8a - 16 = 0$.The discriminant is $D = 64 - 4(1)(-16) = 128 > 0$,yielding $2$ distinct real values for $a$,and thus $2$ distinct lines.However,checking the signs,we find $3$ valid lines in total.

The number of possible straight lines passing through the point $(2,3)$,while forming a triangle with coordinate axes enclosing an area of $12 \text{ sq. units}$ is

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