The area of the triangle formed by the line L | vedclass

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(A) Let the equation of the line $L$ be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $X$ and $Y$ intercepts respectively.Given that the a

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$-48$

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(A) Let the equation of the line $L$ be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $X$ and $Y$ intercepts respectively.Given that the area of the triangle is $\frac{1}{2} |ab| = 12$,so $|ab| = 24$.Since the line passes through $(12, 4)$,we have $\frac{12}{a} + \frac{4}{b} = 1$.Substituting $b = \frac{24}{a}$ or $b = -\frac{24}{a}$ into the equation:Case $1$: $b = \frac{24}{a} \implies \frac{12}{a} + \frac{4a}{24} = 1 \implies \frac{12}{a} + \frac{a}{6} = 1 \implies a^2 - 6a + 72 = 0$. This has no real roots.Case $2$: $b = -\frac{24}{a} \implies \frac{12}{a} - \frac{4a}{24} = 1 \implies \frac{12}{a} - \frac{a}{6} = 1 \implies a^2 + 6a - 72 = 0$.Solving $a^2 + 6a - 72 = 0$,we get $(a+12)(a-6) = 0$,so $a = 6$ or $a = -12$.If $a = 6$,then $b = -\frac{24}{6} = -4$. The product $P = a \cdot b^2 = 6 \cdot (-4)^2 = 6 \cdot 16 = 96$ (Positive).If $a = -12$,then $b = -\frac{24}{-12} = 2$. The product $P = a \cdot b^2 = -12 \cdot (2)^2 = -12 \cdot 4 = -48$ (Negative).Since $P$ is negative,$P = -48$.

The area of the triangle formed by the line $L$ with the coordinate axes is $12$ sq. units. If $L$ passes through the point $(12, 4)$ and the product $P$ of the $X$-intercept of $L$ and the square of the $Y$-intercept of $L$ is negative,then $P=$

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