Straight Line Question 212
Question: The indenter of a triangle with vertices (7, 1), (-1, 5) and $ (3+2\sqrt{3},3+4\sqrt{3}) $ is
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
Correct Answer: A $ \because AB=BC=CA=4\sqrt{5}, $ i.e., given triangle is equilateral. (Incentre of triangle are same as the centroid 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}} ) $
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Answer:
Solution:
BETA
