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  • Consider a family of straight lines <img alt="\mathrm{(x+y)+\lambda(2 x-y+1)=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28x&plus;y%29&plus;%5Clambda%282%20x-y&amp

Consider a family of straight lines \mathrm{(x+y)+\lambda(2 x-y+1)=0}.Equation of the straight line belonging to this family that is farthest form (1,-3) is

Option: 1

13 y+6 x=7


Option: 2

15 y+6 x=7


Option: 3

13 y-6 x=7


Option: 4

15 y-6 x=7


Answers (1)

best_answer

Given family is concurrent at \mathrm{{ }^{\prime} P^{\prime}\left(-\frac{1}{3}, \frac{1}{3}\right)}. If \mathrm{Q=(1,-3)} then \mathrm{m_{P Q}} \mathrm{=\frac{-3-\frac{1}{3}}{1+\frac{1}{3}}=-\frac{5}{2}}.

Now member of the family that is farthest from 'Q'  will have it's slope as \mathrm{\frac{2}{5}}.

\mathrm{\Rightarrow \frac{2}{5}=-\frac{(1+2 \lambda)}{(1-\lambda)} \Rightarrow \lambda=-\frac{7}{8}}
Thus equation of required line is
\mathrm{(x+y)-\frac{7}{8}(2 x-y+1)=0}

I.e. \mathrm{15 y-6 x-7=0}
Hence (D) is the correct answer.

Posted by

shivangi.shekhar

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