Question

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

Answer 1

The given line is $2x+3y=12,$ which can be written as

$\frac{x}{6}+\frac{y}{4}=1...\left(i\right)$

So, the coordinates of the points A and B are $(6,0)$ and $(0,4)$ respectively.

The equation of the line perpendicular to line (i) is

$\frac{x}{4}-\frac{y}{6}+\lambda =0$

This line passes through the point $(5,5)$

$\therefore \frac{5}{4}-\frac{5}{6}+\lambda =0$

$\Rightarrow \lambda =-\frac{5}{12}$

Now, substituting the value of $\lambda$ in $\frac{x}{4}-\frac{y}{6}+\lambda =0,$ we get:

$\frac{x}{4}-\frac{y}{6}-\frac{5}{12}=0$

$\Rightarrow \frac{x}{\frac{5}{2}}-\frac{y}{\frac{5}{2}}=1...\left(ii\right)$

Thus, the coordinates of intersection of line (i) with the x-axis is C $(\frac{5}{3},0)$

To find the coordinates of E, let us write down equations (i) and (ii) in the following manner:

$2x+3y-12=0...\left(iii\right)$

$3x-2y-5=0...\left(iv\right)$

Solving (iii) and (iv) by cross multiplication, we get:

$\frac{x}{-15-24}=\frac{y}{-36+10}=\frac{1}{-4-9}$

$\Rightarrow x=3,y=2$

Thus, the coordinates of E are $(3,2)$ From the figure,

$EC=\sqrt{{(\frac{5}{3}-3)}^2+(0-2{)}^2}=\frac{2\sqrt{13}}{3}$

$EA=\sqrt{(6-3{)}^2+(0-2{)}^2}=\sqrt{13}$

Now,

Area $\left(OCEB\right)$ = Area $\left(\Delta OAB\right)$ - Area $\left(\Delta CAE\right)$

$\Rightarrow Area\left(OCEB\right)=\frac{1}{2}\times 6\times 4-\frac{1}{2}\times \frac{2\sqrt{13}}{3}\times \sqrt{13}$

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