question_answer

The base of an equilateral triangle is along the line given by 3x + 4y = 9. If a vertex of the triangle is (1, 2), then the length of a side of the triangle is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

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\[(a)=\frac{4\sqrt{3}}{5}\]