If the equation of base of an equilateral tri | vedclass

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Question

(a) $AD = \left| {\,\frac{{ - 2 - 2 - 1}}{{\sqrt {{{(2)}^2} + {{( - 1)}^2}} }}\,} \right| = \left| {\,\frac{{ - 5}}{{\sqrt 5 }}\,} \right| = \sqrt 5 $

Vedclass · Accepted Answer

$\sqrt {\frac{{20}}{3}} $

Vedclass · Answer

(a) $AD = \left| {\,\frac{{ - 2 - 2 - 1}}{{\sqrt {{{(2)}^2} + {{( - 1)}^2}} }}\,} ight| = \left| {\,\frac{{ - 5}}{{\sqrt 5 }}\,} ight| = \sqrt 5 $

$	an {60^o}$ $ = \frac{{AD}}{{BD}} \Rightarrow \sqrt 3 = \frac{{\sqrt 5 }}{{BD}}$==> $BD = \sqrt {\frac{5}{3}} $

$	herefore $ $BC = 2BD = 2\sqrt {\frac{5}{3}} = \sqrt {\frac{{20}}{3}} $.

If the equation of base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$, then the length of the side of the triangle is

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