The distance between the points of concurrenc | vedclass

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(B) The given families of lines are: $(x+y+1) + \lambda(5y-3) = 0$  $(i)$ $(2x-3y+3) + \mu(5x+6) = 0$  $(ii)$ For the first family,the point of concur

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$\frac{2 \sqrt{2}}{5}$

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(B) The given families of lines are: $(x+y+1) + \lambda(5y-3) = 0$  $(i)$ $(2x-3y+3) + \mu(5x+6) = 0$  $(ii)$ For the first family,the point of concurrency is the intersection of $x+y+1=0$ and $5y-3=0$. From $5y-3=0$,we get $y = \frac{3}{5}$. Substituting in $x+y+1=0$,we get $x = -\frac{3}{5} - 1 = -\frac{8}{5}$. So,the first point is $P_1 = \left(-\frac{8}{5}, \frac{3}{5}\right)$. For the second family,the point of concurrency is the intersection of $2x-3y+3=0$ and $5x+6=0$. From $5x+6=0$,we get $x = -\frac{6}{5}$. Substituting in $2x-3y+3=0$,we get $3y = 2(-\frac{6}{5}) + 3 = -\frac{12}{5} + 3 = \frac{3}{5}$,so $y = \frac{1}{5}$. So,the second point is $P_2 = \left(-\frac{6}{5}, \frac{1}{5}\right)$. The distance between $P_1$ and $P_2$ is $\sqrt{\left(-\frac{6}{5} - (-\frac{8}{5})\right)^2 + (\frac{1}{5} - \frac{3}{5})^2}$. $= \sqrt{(\frac{2}{5})^2 + (-\frac{2}{5})^2} = \sqrt{\frac{4}{25} + \frac{4}{25}} = \sqrt{\frac{8}{25}} = \frac{2\sqrt{2}}{5}$. Thus,option $(b)$ is correct.

The distance between the points of concurrency of the two families of straight lines given by $x+(5 \lambda+1) y+1-3 \lambda=0$ and $(5 \mu+2) x-3 y+3+6 \mu=0$ is

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