JEE Main2014MathematicsStraight LinesActual
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:
Options
- A2 3 15
- B4 3 15
- C4 3 5
- D2 3 5
Correct answer
B. 4 3 15
Step-by-step solution
Shortest distance of a point (x₁, y₁ ) from line a x+b y=c is d= | a x₁+b y₁-c a^2+b^2 | Now shortest distance of P (1,2) from 3 x+4 y=9 is PC =d= | 3(1)+4(2)-9 3^2+4^2 |= 2 5 Given that APB is an equilateral triangle Let ' a ' be its side then PB =a, CB = a 2 Now, In PCB ,( PB )^2=( PC )^2+( CB )^2 (By Pythagoras theorem) a^2= ( 2 5 )^2+ a^2 4 a^2- a^4 4 = 4 25 3 a^2 4 = 4 25 a^2= 16 75 a= 16 75 = 4 5 3 3 3 = 4 3 15