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 , y axis and the $ AB $ at C, D and E respectively. If O is the origin of coordinates, then the area of $ OCEB $ is [IIT 1976]
Options:
A) $ 23 $ sq. units
B) $ \frac{23}{2}sq. $ units
C) $ \frac{23}{3} $ sq. units
D)None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- 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 $ ( \frac{5}{3},0 ) $ . Similarly the coordinates of E are (3, 2) by solving the line AB and (i). Thus O(0, 0), $ C( \frac{5}{3},0 ) $ , $ 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.
BETA
