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Question

The line 2 x+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.

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Solution

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

x6+y4=1 ...(i)

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

x4y6+λ=0

This line passes through the point (5, 5)

5456+λ=0

λ=512

Now, substituting the value of λ in x4y6+λ=0, we get:

x4y6512=0

x52y52=1 ...(ii)

Thus, the coordinates of intersection of line (i) with the x-axis is C (53,0)

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

2x+3y12=0 ...(iii)

3x2y5=0 ...(iv)

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

x1524=y36+10=149

x=3, y=2

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

EC=(533)2+(02)2=2133

EA=(63)2+(02)2=13

Now,

Area (OCEB) = Area (Δ OAB) - Area (ΔCAE)

Area(OCEB)=12×6×412×2133×13

=233 sq.units


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