Coordinate Geometry Question 8
Question: The following points A (2a, 4a), B(2a, 6a) and C $ (2a+\sqrt{3}a,5a) $ , $ (a>0) $ are the vertices of
Options:
A) An acute angled triangle
B) A right angled triangle
C) An isosceles triangle
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ AB=\sqrt{{{(2a-2a)}^{2}}+{{(4a-6a)}^{2}}}=2a $
$ BC=\sqrt{{{(\sqrt{3}a)}^{2}}+a^{2}}=2a $
$ CA=\sqrt{{{(\sqrt{3}a)}^{2}}+{{(-a)}^{2}}}=2a $ Since $ AB=BC=CA, $ hence triangle is equilateral. Therefore, it is an acute angled triangle.
BETA
