Question

The line $2x+3y=12$ meets the coordinates axes at $A$ and $B$ respectively. The line through $\left(5,5\right)$ perpendicular to $AB$ meets the coordinate axes and the line $AB$ at $C, \, D$ and $E$ respectively. If $O$ is the origin, then the area (in sq. units) of the figure $OCEB$ is equal to

• A. $\frac{13}{3}$
• B. $\frac{23}{3}$
• C. $11$
• D. $7$

Answer 1

Answer: (B)

The equation of a line passing through $P\left(5,5\right)$ and perpendicular to $2x+3y=12$ is
$y-5=\frac{3}{2}\left(x - 5\right)$
or, $3x-2y-5=0$
The coordinates of the point of intersection of $2x+3y-12=0$ and $3x-2y-5=0$ are $E\left(3,2\right).$
The line $3x-2y-5=0$ meets $x$ -axis at $C\left(\frac{5}{3} , 0\right).$
Now,
$AreaoffigureOCEB$
$=Areaof\Delta AOB-Areaof\Delta ACE$
$=\frac{1}{2}\left(O A \times O B\right)-\frac{1}{2}\left(A C \times E L\right)$
$=\frac{1}{2}\times 6\times 4-\frac{1}{2}\times \frac{13}{3}\times 2$
$=\frac{23}{3}sq.units$