A) \[\frac{2\sqrt{3}}{15}\]
B) \[\frac{4\sqrt{3}}{15}\]
C) \[\frac{4\sqrt{3}}{5}\]
D) \[\frac{2\sqrt{3}}{5}\]
Correct Answer: B
Solution :
Shortest distance of a point \[({{x}_{1}},{{y}_{1}})\]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 P (1, 2) from 3x + 4y = 9\[PC=d=\left| \frac{3(1)+4(2)-9}{\sqrt{{{3}^{2}}+{{4}^{2}}}} \right|=\frac{2}{5}\] Given that DAPB is an equilateral triangle Let 'a' be its side then \[PB=a,CB=\frac{a}{2}\] Now, In \[\Delta PCB,{{(PB)}^{2}}={{(PC)}^{2}}+{{(CB)}^{2}}\](By Pythagoras theorem) \[{{a}^{2}}={{\left( \frac{2}{5} \right)}^{2}}+\frac{{{a}^{2}}}{4}\] \[{{a}^{2}}-\frac{{{a}^{4}}}{4}=\frac{4}{25}\Rightarrow \frac{3{{a}^{2}}}{4}=\frac{4}{25}\] \[{{a}^{2}}=\frac{16}{75}\Rightarrow a=\sqrt{\frac{16}{75}}=\frac{4}{5\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{15}\] \[\therefore \]Length of Equilateral triangle\[(a)=\frac{4\sqrt{3}}{5}\]You need to login to perform this action.
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