Coordinate Geometry Question 156
Question: The incentre of triangle formed by the lines $ x=0, $
$ y=0 $ and $ 3x+4y=12 $ is
[RPET 1990]
Options:
A) $ ( \frac{1}{2},\frac{1}{2} ) $
B) (1, 1)
C) $ ( 1,\frac{1}{2} ) $
D) $ ( \frac{11}{2},1 ) $
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ a=BC=5,b=AC=4,c=AB=3 $
Hence incentre is $ ( \frac{0+0+3\times 4}{5+4+3},\frac{0+4\times 3+0}{5+4+3} ) $ = (1, 1).
BETA
