If the equation of the base of an equilateral | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the vertex be $A(-1, 2)$ and the base be $BC$ with equation $2x - y - 1 = 0$.The altitude $AD$ from vertex $A$ to the base $BC$ is the perpendicular distance from $A$ to the line $2x - y - 1 = 0$.$AD = \left| \frac{2(-1) - (2) - 1}{\sqrt{2^2 + (-1)^2}} ight| = \left| \frac{-2 - 2 - 1}{\sqrt{5}} ight| = \frac{5}{\sqrt{5}} = \sqrt{5}$.In an equilateral triangle,the altitude $AD$ is related to the side length $s$ by $AD = s \sin(60^o) = s \frac{\sqrt{3}}{2}$.Therefore,$\sqrt{5} = s \frac{\sqrt{3}}{2}$ $\Rightarrow s = \frac{2\sqrt{5}}{\sqrt{3}} = \sqrt{\frac{4 	imes 5}{3}} = \sqrt{\frac{20}{3}}$.

If the equation of the 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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