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

Engineering

Mathematics

Area of a Triangle

Locus of a Point

Various Form of a Straight Line

Question

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

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

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

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

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

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Equation of line ED is 3x – 2y = 5
∴   pt E is (3, 2) C(5/3, 0)
now area of OCEB = AR(ΔOBC+ΔBEC)
$ΔOBC = \frac{1}{2} \times 4 \times \frac{5}{3} = \frac{10}{3}$ ....(1)
$ΔBEC = \frac{1}{2} (BE \times CE)$
$= \frac{1}{2} \times \sqrt{13} \frac{\sqrt{25 \times 13}}{3} = \frac{5 \times 13}{2 \times 3}$
$\frac{1}{2} \times \sqrt{13} \times \frac{2}{3} \sqrt{13} = \frac{13}{3}$        ....(2)
   Area OCEB $= \frac{10}{3} + \frac{13}{3} = \frac{23}{3} sq. units$