Line 2x + 3y = 12 meets x -axis at A and y -axis at B . line through (5, 5) perpendicular to AB meet

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9.Straight Line

hard

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 , $y$ axis and the $AB$ at $C,\,D$ and $E$ respectively. If $O$ is the origin of coordinates, then the area of $OCEB$ is

$23\, sq. units$

$\frac{{23}}{2}sq. units$

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

None of these

(IIT-1976)

Solution

(c) Here $ O$ is the point $(0,\,0)$. The line $2x + 3y = 12$ meets the $y$-axis at $B$ and so $B$ is the point $(0,4)$. The equation of any line perpendicular to the line $2x + 3y = 12$ and passes through $(5, 5)$ is $3x – 2y = 5$……$(i)$

The line $(i)$ meets the $x$-axis at $C$ and so co-ordinates of $C$ are$\left( {\frac{5}{3},\,0} \right).$Similarly the coordinates of $E$ are $(3, 2)$ by solving the line $AB$ and $(i)$.

Thus $O\,(0, 0)$, $C\left( {\frac{5}{3},0} \right)$, $E(3,\,2)$ and $B \,(0, 4)$. Now the area of figure $OCEB = $ area of $\Delta OCE$ + area of $\Delta OEB = \frac{{23}}{3}sq.$ units.

Standard 11

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