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(A) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the intercepts on the axes.The area of the triangle formed w

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$\frac{x}{4} + \frac{y}{3} = \pm 1$

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(A) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the intercepts on the axes.The area of the triangle formed with the axes is given by $\frac{1}{2} |ab| = 6$,which implies $|ab| = 12$ $(i)$.The length of the hypotenuse is given by $\sqrt{a^2 + b^2} = 5$,which implies $a^2 + b^2 = 25$ $(ii)$.From $(i)$,$b = \frac{12}{a}$. Substituting this into $(ii)$:$a^2 + (\frac{12}{a})^2 = 25$$a^4 - 25a^2 + 144 = 0$$(a^2 - 16)(a^2 - 9) = 0$So,$a^2 = 16$ or $a^2 = 9$,which gives $a = \pm 4$ or $a = \pm 3$.If $a = \pm 4$,then $b = \pm 3$. If $a = \pm 3$,then $b = \pm 4$.The equations of the lines are $\frac{x}{\pm 4} + \frac{y}{\pm 3} = 1$ or $\frac{x}{\pm 3} + \frac{y}{\pm 4} = 1$,which can be written as $\frac{x}{4} + \frac{y}{3} = \pm 1$ (and other variations).Checking the options,$\frac{x}{4} + \frac{y}{3} = \pm 1$ satisfies the given conditions.

The equation of the line which makes a right-angled triangle with the axes whose area is $6$ sq. units and whose hypotenuse is $5$ units,is

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