A) \[\frac{x}{4}+\frac{y}{3}=\pm \ 1\]
B) \[\frac{x}{4}-\frac{y}{3}=\pm \ 3\]
C) \[\frac{x}{6}+\frac{y}{1}=\pm \ 1\]
D) \[\frac{x}{1}-\frac{y}{6}=\pm \ 1\]
Correct Answer: A
Solution :
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.
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