The equation of the base of an equilateral tr | vedclass

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(C) Since the vertex $A(2,3)$ does not lie on the line $12x+5y-65=0$,the perpendicular distance from $A$ to the base $BC$ is the altitude $AD$ of the

Vedclass · Accepted Answer

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

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(C) Since the vertex $A(2,3)$ does not lie on the line $12x+5y-65=0$,the perpendicular distance from $A$ to the base $BC$ is the altitude $AD$ of the equilateral triangle $	riangle ABC$.The length of the altitude $AD$ is given by the formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $ax+by+c=0$:$AD = \left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}ight|$$AD = \left|\frac{12(2)+5(3)-65}{\sqrt{12^2+5^2}}ight| = \left|\frac{24+15-65}{\sqrt{144+25}}ight| = \left|\frac{39-65}{\sqrt{169}}ight| = \left|\frac{-26}{13}ight| = 2$In an equilateral triangle with side length $s$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}s$.Therefore,$s = \frac{2}{\sqrt{3}} 	imes h$.Substituting $h = AD = 2$:$s = \frac{2}{\sqrt{3}} 	imes 2 = \frac{4}{\sqrt{3}}$.

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$,then the length of the side is

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