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The line \mathrm{2 x+3 y=12} meets the \mathrm{x} - axis at \mathrm{A} and the \mathrm{y}- axis at \mathrm{B}. The line through (5,5) perpendicular to \mathrm{AB} meets the \mathrm{x}-axis, \mathrm{y}-axis & the line \mathrm{AB} at \mathrm{C}, \mathrm{D},\mathrm{E} respectively. If \mathrm{O} is the origin, then the area of the OCEB is :  

Option: 1

\frac{20}{3} \mathrm{sq. units}


Option: 2

\frac{23}{3} \mathrm{sq. units}


Option: 3

\frac{26}{3} \mathrm{sq. units}


Option: 4

\frac{5 \sqrt{52}}{9}\mathrm{sq. units}


Answers (1)

best_answer

Equation of line \mathrm{ED}$ is $3 \mathrm{x}-2 \mathrm{y}=5

\therefore \mathrm{pt} \: \mathrm{E}$ is $(3,2) \mathrm{C}(5 / 3,0)
\text{now area of}\: \mathrm{OCEB}=\mathrm{AR}\left(\Delta \mathrm{OBC}+{ }{\Delta} \mathrm{BEC}\right)

\triangle \mathrm{OBC}=\frac{1}{2} \times 4 \times \frac{5}{3}=\frac{10}{3}\quad \ldots{(1)}
\Delta \mathrm{BEC}=\frac{1}{2}(\mathrm{BE} \times \mathrm{CE})

              =\frac{1}{2} \times \sqrt{13} \frac{\sqrt{25 \times 13}}{3}=\frac{5 \times 13}{2 \times 3}

\frac{1}{2} \times \sqrt{13} \times \frac{2}{3} \sqrt{13}=\frac{13}{3} \quad \ldots(2)
\mathrm{Area \: OCEB =\frac{10}{3}+\frac{13}{3}=\frac{23}{3} \: sq. units}



 

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Anam Khan

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