The product of the perpendiculars from (1,1) to the pair of straight lines 2x^2 + 6xy + 3y^2 = 0 is

And:- 11/(37)^0.5

I have been trying this question 5 or 6 times now and still bot getting the answer... Can anyone please help me out ? Thanks in advance :)

Dear Gautham

let y=m₁x  and y=m₂x  are two lines  and p₁ and p₂ are the length of perpendicular from (1,1) on these line respectively

the m₁ +m₂ =-6/3 =-2

m₁.m₂ =2/3

 p₁ =(1*1 -1*m₁)/√(1+m₁²)

      =(1-m₁)/√(1+m₁²)

similerly

p₂ =(1*1 -1*m₁)/√(1+m₁²)

      =(1-m₂)/√(1+m₂²)

so p1 .p2 ={(1-m₁)/√(1+m₁²)}*{(1-m₂)/√(1+m₂²)}

               ={1-(m₁₊ m_{2) +}m₁m_2}/√{1+(m₁m_{2)² +(}m_{1 +}m_{2 )² -2}m₁m_2}

               ={1+2+2/3}/√{1+ 4/9 +4 -4/3_}

               =11/√37

Thanks so much :) A lot faster method than what I thought was the method :)