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(B) Let the vertex be $A(1, 2)$ and the base $BC$ lie on the line $2x - y - 1 = 0$. The altitude $AD$ from vertex $A$ to the base $BC$ is the perpendi

Vedclass · Accepted Answer

$\frac{2}{\sqrt{15}}$

Vedclass · Answer

(B) Let the vertex be $A(1, 2)$ and the base $BC$ lie on the line $2x - y - 1 = 0$. The altitude $AD$ from vertex $A$ to the base $BC$ is the perpendicular distance from $(1, 2)$ to the line $2x - y - 1 = 0$. $AD = \left| \frac{2(1) - (2) - 1}{\sqrt{2^2 + (-1)^2}} \right| = \left| \frac{2 - 2 - 1}{\sqrt{5}} \right| = \frac{1}{\sqrt{5}}$. In an equilateral triangle,the altitude $AD$ is related to the side length $s$ by $AD = s \sin 60^{\circ} = s \frac{\sqrt{3}}{2}$. Therefore,$\frac{1}{\sqrt{5}} = s \frac{\sqrt{3}}{2}$. $s = \frac{2}{\sqrt{5} \cdot \sqrt{3}} = \frac{2}{\sqrt{15}}$.

The base of an equilateral triangle is represented by the equation $2x - y - 1 = 0$ and its vertex is $(1, 2)$. Then,the length (in units) of the side of the triangle is:

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