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If \mathrm{m}_{1}, \mathrm{m}_{2} are the roots of the equation \mathrm{x}^{2}-\mathrm{ax}-\mathrm{a}-1=0, then the area of the triangle formed by the three straight lines \mathrm{y=m_{1} x, y=m_{2} x\: and \: y=a(a \neq-1)} is

Option: 1

\mathrm{\frac{a^{2}(a+2)}{2(a+1)},\: if\: a>-1}


Option: 2

\mathrm{-\frac{a^{2}(a+2)}{2(a+1)},\: if \: a<-1\: }


Option: 3

\mathrm{\frac{a^{2}(a+2)}{2(a+1)}, \: if -2<a<-1}


Option: 4

\mathrm{-\frac{a^{2}(a+2)}{2(a+1)}\, if \, a<-2}


Answers (1)

best_answer

Vertices of the given triangle are \mathrm{(0,0)\left(\frac{\mathrm{a}}{\mathrm{m}_{1}}, \mathrm{a}\right) and \left(\frac{\mathrm{a}}{\mathrm{m}_{2}}, \mathrm{a}\right)}  so that the area of the triangles is equal to \mathrm{\frac{\mathrm{a}^{2}}{2}\left|\frac{\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right)}{\mathrm{m}_{1} \mathrm{~m}_{2}}\right|}

since, \mathrm{\mathrm{m}_{1}, \mathrm{m}_{2}} re the roots of \mathrm{\mathrm{x}^{2}-\mathrm{ax}-\mathrm{a}-1=0}, so \mathrm{\mathrm{m}_{1}+\mathrm{m}_{2}=\mathrm{a}, \mathrm{m}_{1} \mathrm{~m}_{2}=-(\mathrm{a}+1)}
\mathrm{\Rightarrow\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)^{2}=\mathrm{a}^{2}+4(\mathrm{a}+1)=(\mathrm{a}+2)^{2}}
\mathrm{ \Rightarrow\left|\mathrm{m}_{1}-\mathrm{m}_{2}\right|=|(\mathrm{a}+2)|}

so, the required area is \mathrm{ \Delta=\frac{a^{2}}{2}\left|\frac{a+2}{-(a+1)}\right|=\frac{a^{2}}{2}\left|\frac{a+2}{a+1}\right|}
since the area of \mathrm{ \Delta} is a positive quantity.
\mathrm{\Delta=\frac{a^{2}(a+2)}{2(a+1)}, \quad \text { if } a>-1 \text { or } a<-2}
\mathrm{\text { and } \Delta=-\frac{a^{2}(a+2)}{2(a+1)}, \text { if }-2<a<-1 }.
 

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seema garhwal

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