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(B) The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

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$\frac{4\sqrt{3}}{15}$

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(B) The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.For the point $P(1, 2)$ and the line $3x + 4y - 9 = 0$,the altitude $h$ of the equilateral triangle is:$h = \frac{|3(1) + 4(2) - 9|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 9|}{\sqrt{9 + 16}} = \frac{2}{5}$.In an equilateral triangle with side length $a$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}a$.Equating the two expressions for $h$:$\frac{\sqrt{3}}{2}a = \frac{2}{5}$Solving for $a$:$a = \frac{2}{5} 	imes \frac{2}{\sqrt{3}} = \frac{4}{5\sqrt{3}}$.Rationalizing the denominator:$a = \frac{4}{5\sqrt{3}} 	imes \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{15}$.

The base of an equilateral triangle is along the line given by $3x + 4y = 9$. If a vertex of the triangle is $(1, 2)$,then the length of a side of the triangle is

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