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 x4−y6+λ=0 This line passes through the point (5, 5) ∴ 54−56+λ=0 ⇒ λ=−512 Now, substituting the value of λ in x4−y6+λ=0, we get: x4−y6−512=0 ⇒ x52−y52=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+3y−12=0 ...(iii) 3x−2y−5=0 ...(iv) Solving (iii) and (iv) by cross multiplication, we get: x−15−24=y−36+10=1−4−9 ⇒ x=3, y=2 Thus, the coordinates of E are (3, 2) From the figure, EC=√(53−3)2+(0−2)2=2√133 EA=√(6−3)2+(0−2)2=√13 Now, Area (OCEB) = Area (Δ OAB) - Area (ΔCAE) ⇒ Area(OCEB)=12×6×4−12×2√133×√13 =233 sq.units

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