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

\[\frac{x}{6} + \frac{y}{4} = 1\]         ... (1)
So, the coordinates of the points A and B are (6, 0) and (0, 4), respectively.

The equation of the line perpendicular to line (1) 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:

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

\[ \Rightarrow \frac{x}{\frac{5}{3}} - \frac{y}{\frac{5}{2}} = 1 . . . (2)\]

Thus, the coordinates of intersection of line (1) with the x-axis is \[C \left( \frac{5}{3}, 0 \right)\].

To find the coordinates of E, let us write down equations (1) and (2) in the following manner: 

\[2x + 3y - 12 = 0\]            ... (3) 

\[3x - 2y - 5 = 0\]            .. (4)

Solving (3) and (4) 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, \[EC = \sqrt{\left( \frac{5}{3} - 3 \right)^2 + \left( 0 - 2 \right)^2} = \frac{2\sqrt{13}}{3}\] 

\[EA = \sqrt{\left( 6 - 3 \right)^2 + \left( 0 - 2 \right)^2} = \sqrt{13}\]

Now,

\[\text { Area  }\left( OCEB \right) = \text { Area } \left( ∆ OAB \right) - \text { Area } \left( ∆ CAE \right)\]

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

\[ = \frac{23}{3} \text { sq units }\]