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

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Aicharya · Accepted Answer

Shortest distance of a point $\left(x_1, y_1\right)$ from line $a x+b y=c$ is $d=\left|\frac{a x_1+b y_1-c}{\sqrt{a^2+b^2}}\right|$ Now shortest distance of $\mathrm{P}(1,2)$ from $3 x$ $+4 y=9$ is $ \mathrm{PC}=d=\left|\frac{3(1)+4(2)-9}{\sqrt{3^2+4^2}}\right|=\frac{2}{5} $ Given that $\triangle \mathrm{APB}$ is an equilateral triangle Let ' $a$ ' be its side then $\mathrm{PB}=a, \mathrm{CB}=\frac{a}{2}$ Now, In $\Delta \mathrm{PCB},(\mathrm{PB})^2=(\mathrm{PC})^2+(\mathrm{CB})^2$ (By Pythagoras theorem) $ a^2=\left(\frac{2}{5}\right)^2+\frac{a^2}{4} $ $a^2-\frac{a^4}{4}=\frac{4}{25} \Rightarrow \frac{3 a^2}{4}=\frac{4}{25}$ $a^2=\frac{16}{75} \Rightarrow a=\sqrt{\frac{16}{75}}=\frac{4}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4 \sqrt{3}}{15}$

The base of an equilateral triangle is along the line given by $3 x+4 y=9$. If a vertex of the triangle is…

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