Equation of base BC of an equilateral triangle is 3x + 4y = 1 and vertex is (-3,2), then area o

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

A

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

B

$\frac{4}{5\sqrt 3}$

C

$\frac{8\sqrt 3}{75}$

D

$\frac{16\sqrt 3}{25}$

Solution

$\mathrm{CM}=\frac{|-9+8-1|}{5}=\frac{2}{5}$

Now ${\rm{AC}} = \frac{2}{5}\cos ec{60^\circ } = \frac{4}{{5\sqrt 3 }}$

$\therefore {\mathop{\rm area}\nolimits} (\Delta {\rm{ABC}}) = \frac{{\sqrt 3 }}{4} \cdot \frac{{16}}{{25 \times 3}} = \frac{{4\sqrt 3 }}{{75}}$

Standard 11

Mathematics

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