question_answer

If a straight line\[ax+by=c\]meets\[x-\]axis at\[A\]and\[y-\]axis at\[B\]. Then area of triangle \[OAB\] where\[O\]is the point of intersection of coordinates axes, is

A) \[\frac{{{c}^{2}}}{ab}\]                                 

B) \[\frac{{{c}^{2}}}{2ab}\]  

C) \[\frac{{{c}^{2}}}{{{a}^{2}}b}\]                  

D)        \[\frac{c}{{{b}^{2}}a}\]

Correct Answer: B

Solution :

The equation is:\[ax+by=c\] Putting,\[x=0,\,\,by=c\Rightarrow y=\frac{c}{b}\] Putting,\[y=0,\,\,ax=c\Rightarrow x=\frac{c}{a}\] Clearly,\[OA=\frac{c}{a},\,\,OB=\frac{c}{b}\] \[\therefore \]\[ar(OAB)=\frac{1}{2}OA\times AB\] \[=\frac{1}{2}\times \frac{c}{a}\times \frac{c}{b}=\frac{{{c}^{2}}}{2ab}\]