Coordinate Geometry Question 35
Question: The incentre of a triangle with vertices (7, 1) (-1, 5) and $ (3+2\sqrt{3},3+4\sqrt{3}) $ is
[J & K 2005]
Options:
A) $ ( 3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}} ) $
B) $ ( 1+\frac{2}{3\sqrt{3}},1+\frac{4}{3\sqrt{3}} ) $
C) (7, 1)
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \becauseAB=BC=CA=4\sqrt{5} $ , i.e., given triangle is equilateral. (In centre of a triangle are same as the centriod when triangle is equilateral)
Hence, incentre = $ ( \frac{7-1+3+2\sqrt{3}}{3},\frac{1+5+3+4\sqrt{3}}{3} ) $
$ =( 3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}} ) $ .
BETA
