The equation of the base of an equilateral tr | vedclass

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(C) Let $	riangle ABC$ be the equilateral triangle with base $BC$ given by $x+y-2=0$ and vertex $A$ as $(2,-1)$.The length of the altitude $h$ from v

Vedclass · Accepted Answer

$\sqrt{2/3}$

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(C) Let $	riangle ABC$ be the equilateral triangle with base $BC$ given by $x+y-2=0$ and vertex $A$ as $(2,-1)$.The length of the altitude $h$ from vertex $A$ to the base $BC$ is the perpendicular distance from $(2,-1)$ to $x+y-2=0$.$h = \frac{|(1)(2) + (1)(-1) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|2 - 1 - 2|}{\sqrt{2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.In an equilateral triangle with side length $s$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}s$.Therefore,$\frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}s$.$s = \frac{2}{\sqrt{2} 	imes \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.

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

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