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 & the line AB at C,D,E respectively . if O is the origin , then the area of the OCEB is:

Answer 1

$2x+3y=12$

$\frac{x}{6}+\frac{y}{4}=1$

Using point form,

$A\equiv (6,0);B=(0,4)$

also, slope of line $AB=\frac{-2}{3}$

So, slope of perpendicular $=\frac{-1}{m}=\frac{+3}{2}$

So, equation will be,

$y-5=\frac{3}{2}(x-5)3x-2y=5\frac{3x}{5}-\frac{2y}{5}=1$

Using point form, $C\equiv (\frac{5}{3},0);D\equiv (0,\frac{-5}{2})$

and solving with $2x+3y=12,$ we get $E\equiv (3,2).$

So, $O\equiv (0,0)C\equiv (\frac{5}{3},0)E\equiv (3,2)B\equiv (0,4)$

$Ar\left(OCEB\right)=|Ar\left(OCE\right)|+|Ar\left(OEB\right)|$

$Ar\left(OCEB\right)=\frac{1}{2}\times \frac{5}{3}+\frac{1}{2}\times 12=\frac{41}{6}$