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

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(B) Let the vertex be $A(2, -1)$ and the base be $BC$ with the equation $x + 2y - 1 = 0$.The length of the altitude $AD$ from vertex $A$ to the base $

Vedclass · Accepted Answer

$2/\sqrt{15}$

Vedclass · Answer

(B) Let the vertex be $A(2, -1)$ and the base be $BC$ with the equation $x + 2y - 1 = 0$.The length of the altitude $AD$ from vertex $A$ to the base $BC$ is the perpendicular distance from point $A$ to the line $x + 2y - 1 = 0$.$AD = \frac{|(1)(2) + (2)(-1) - 1|}{\sqrt{1^2 + 2^2}} = \frac{|2 - 2 - 1|}{\sqrt{5}} = \frac{1}{\sqrt{5}}$.In an equilateral triangle,the altitude $AD$ is related to the side length $s$ by the formula $AD = s \frac{\sqrt{3}}{2}$.Therefore,$s \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{5}}$.$s = \frac{2}{\sqrt{3} \cdot \sqrt{5}} = \frac{2}{\sqrt{15}}$.

The vertex of an equilateral triangle is $(2, -1)$ and the equation of its base is $x + 2y - 1 = 0$. The length of its side is

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