Coordinate Geometry Question 28
Question: The points (3a, 0), (0, 3b) and (a, 2b) are
[MP PET 1982]
Options:
A) Vertices of an equilateral triangle
B) Vertices of an isosceles triangle
C) Vertices of a right angled isosceles triangle
D) Collinear
Show Answer
Answer:
Correct Answer: D
Solution:
$ l_1=\sqrt{{{(3a)}^{2}}+{{(3b)}^{2}}}=3\sqrt{a^{2}+b^{2}} $
$ l_2=\sqrt{a^{2}+b^{2}}=\sqrt{a^{2}+b^{2}} $
$ l_3=\sqrt{{{(2a)}^{2}}+{{(2b)}^{2}}}=2\sqrt{a^{2}+b^{2}} $
$ \Rightarrow l_1=l_2+l_3 $
Hence the points are collinear.
BETA
