Mathematics - Point Question with Solution | TestHub

Shortest distance of a point $\left(x_1,y_1\right)$ from line $ax+by=c$ 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 $3x$ $+4y=9$ is $$\mathrm{PC}=d=\left|\frac{3(1)+4(2)-9}{\sqrt{3^2+4^2}}\right|=\frac{2}{5}$$ Given that $△\mathrm{APB}$ 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) $$a^2={\left(\frac{2}{5}\right)}^2+\frac{a^2}{4}$$ $a^2-\frac{a^4}{4}=\frac{4}{25}\Rightarrow \frac{3a^2}{4}=\frac{4}{25}$ $a^2=\frac{16}{75}\Rightarrow a=\sqrt{\frac{16}{75}}=\frac{4}{5\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{15}$