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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 :)

7 years ago

147 Points

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+ m2) +m1m2}/√{1+(m1m2)2 +(m1 +m2 )2 -2m1m2}

={1+2+2/3}/√{1+ 4/9 +4 -4/3}

=11/√37

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Regards,

7 years ago
Gautham Warrier
4 Points
Thanks so much :) A lot faster method than what I thought was the method :)

7 years ago
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