Straight Line Question 117

Question: The equation of the base of an equilateral triangle is $ x+y=2 $ and the vertex is (2, -1). The length of the side of the triangle is [IIT 1973, 83, MP PET 1995; RPET 1999, 2000]

Options:

A) $ \sqrt{3/2} $

B) $ \sqrt{2} $

C) $ \sqrt{2/3} $

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

  • Let p be the length of the perpendicular from the vertex (2, ?1) to the base $ x+y=2 $ . Then $ p=| \frac{2-1-2}{\sqrt{1^{2}+1^{2}}} |=\frac{1}{\sqrt{2}} $ If ‘a’ be the length of the side of triangle, then $ p=a\sin 60^{o}\Rightarrow \frac{1}{\sqrt{2}}=\frac{a\sqrt{3}}{2}\Rightarrow a=\sqrt{\frac{2}{3}} $ .



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