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A straight line \mathrm{L} with positive slope passes through the points \mathrm{(4,-4)} and cuts the co-ordinate axes in fourth quadrants at points \mathrm{A \: and \: B}. As\mathrm{L} varies, the absolute minimum value of \mathrm{\mathrm{OA}+\mathrm{OB}}  is ( 0 is origin)
 

Option: 1

0


Option: 2

16


Option: 3

8


Option: 4

18


Answers (1)

Equation of line is \mathrm{y+4=m(x-4), as \: m>0}

\mathrm{ \therefore \quad \mathrm{mx}-\mathrm{y}=4 \mathrm{~m}+4 }

\mathrm{ \frac{x}{\frac{4 m+4}{m}}+\frac{y}{-(4 m+4)}=1 }

\mathrm{\therefore \quad O A+O B=\frac{4 m+4}{m}+4 m+4 }

\mathrm{ =4 m+\frac{4}{m}+8 \geq 16 }

Hence option 2 is correct.





 

Posted by

Kshitij

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