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Equation of a line given some parametrized points on it and an area

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My kid has this question for homework:

A line passing through the point $(6,6)$ crosses the line $y=2x$ at the point $A$, and the $x$-axis at the point $B$ (the $x$-coordinate of $B$ is positive). Find the line's equation if the area of the triangle $ABO$ is $48$ ($O$ is the origin).

If we set $A=(a,2a)$ and $B=(b,0)$, then the area restriction implies that $ab=48$. But beyond that, my kid is at a loss as to how to proceed. Can you help, please?

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Because $(6,6)$ is on the line, we have the slope $\frac{6-0}{6-b}=\frac{6-2a}{6-a}$, which simplifies to $3a+3b=ab$. We already know that's $48$ so $a+b=16, ab=48$, and thus $\lbrace a,b\rbrace=\lbrace4,12\rbrace$. This gives two possibilities for where $A$ and $B$ are; finding the respective equations of the line is then easy using one of the other points on the line.

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