The equation of the base BC of an equilateral | vedclass

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Question

(A) Let the vertex be $A(-3, 2)$ and the base be $BC$ with equation $3x + 4y - 1 = 0$.The altitude $h$ from $A$ to $BC$ is the perpendicular distance

Vedclass · Accepted Answer

$\frac{4\sqrt{3}}{75}$

Vedclass · Answer

(A) Let the vertex be $A(-3, 2)$ and the base be $BC$ with equation $3x + 4y - 1 = 0$.The altitude $h$ from $A$ to $BC$ is the perpendicular distance from $A$ to the line $BC$:$h = \frac{|3(-3) + 4(2) - 1|}{\sqrt{3^2 + 4^2}} = \frac{|-9 + 8 - 1|}{\sqrt{25}} = \frac{|-2|}{5} = \frac{2}{5}$.In an equilateral triangle with side length $s$,the altitude $h = \frac{\sqrt{3}}{2}s$,so $s = \frac{2h}{\sqrt{3}} = \frac{2(2/5)}{\sqrt{3}} = \frac{4}{5\sqrt{3}}$.The area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$.Area $= \frac{\sqrt{3}}{4} 	imes \left(\frac{4}{5\sqrt{3}}ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{16}{25 	imes 3} = \frac{4\sqrt{3}}{75}$.

The equation of the base $BC$ of an equilateral triangle is $3x + 4y = 1$ and the vertex $A$ is $(-3, 2)$. Then the area of the triangle is-

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