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Find the vertices of the two right angled triangles each having area $18$ and such that the point $(2,4)$ lies on the hypotenuse and the other two sides are formed by the $x$ and $y$ axes.

My work:
Since, the triangle is formed by the $x$ and $y$ axes, the product of the intercepts is $36$.
So,we have to find $2$ lines which has product of intercepts equal to $36$ and which passes through $(2,4)$.From here, I can solve the problem by trial and error,but I want to do this sum without any trial-error method.
So, I form $2$ equations of line with it. But, I cannot proceed any further. Please help!

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If $A = (a,0)$ and $B = (0,b)$ are the intercepts, we have:

(1) $ab = 36$, as you say.
(2) The line $AB$, given by $\frac{x}{a} + \frac{y}{b} = 1$, also passes through $(2,4)$, so $\frac{2}{a} + \frac{4}{b} = 1$.

From (2), $2b + 4a = ab = 36$, so $b = 18 - 2a$. Now substitute this in (1) to get a quadratic equation in $a$.

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