The vertex of an equilateral triangle is (2,- | vedclass

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Question

(b) $|AD|\, = \left| {\frac{{2 - 2 - 1}}{{\sqrt {{1^2} + {2^2}} }}} ight| = \frac{1}{{\sqrt 5 }}$

$	an 60^\circ = \frac{{AD}}{{BD}}$

==> $

Vedclass · Accepted Answer

$2/\sqrt {15} $

Vedclass · Answer

(b) $|AD|\, = \left| {\frac{{2 - 2 - 1}}{{\sqrt {{1^2} + {2^2}} }}} ight| = \frac{1}{{\sqrt 5 }}$

$	an 60^\circ = \frac{{AD}}{{BD}}$

==> $\sqrt 3 = \frac{{1\,/\sqrt 5 }}{{BD}}$

==> $BD = \frac{1}{{\sqrt {15} }}$

$BC = 2BD = 2\,/\sqrt {15} $.

The vertex of an equilateral triangle is $(2,-1)$ and the equation of its base in $x + 2y = 1$. The length of its sides is

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