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The equation of the line which makes right angled triangle with axes whose area is $6$ sq. units and whose hypotenuse is of $5$ units, is
$\frac{x}{4} + \frac{y}{3} = \pm \;1$
$\frac{x}{4} - \frac{y}{3} = \pm \;3$
$\frac{x}{6} + \frac{y}{1} = \pm \;1$
$\frac{x}{1} - \frac{y}{6} = \pm \;1$
Solution
(a) If the line is $\frac{x}{a} + \frac{y}{b} = 1$, then the intercepts on the axes are $a$ and $b$.
Therefore the area is $\frac{1}{2}|a \times b| = 6 \Rightarrow |ab| = 12$ …..$(i)$
and hypotenuse is $5$, therefore ${a^2} + {b^2} = 25$ …..$(ii)$
On solving $(i)$ and $(ii)$, we get
$a = \pm 4$or $ \pm 3$and $b = \pm 3$or $ \pm 4$
Hence equation of line is $ \pm \frac{x}{4} \pm \frac{y}{3} = 1$or $ \pm \frac{x}{3} \pm \frac{y}{4} = 1$.
Trick: Check with options. Obviously, the line $\frac{x}{4} + \frac{y}{3} = \pm 1$ satisfies both the conditions.
