Given the family of lines, $a(2x + y + 4) + b(x - 2y - 3) = 0$. Among the lines of the family, the n

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Given the family of lines, $a(2x + y + 4) + b(x - 2y - 3) = 0$. Among the lines of the family, the number of lines situated at a distance of $\sqrt{10}$ from the point $M(2, - 3)$ ) is:

Accepted Answer

Sol. [Key Concept: Find the fixed point of the family; the count of lines at distance $\sqrt{10}$ from $M$ depends on $|$fixed point $- M|$ vs $\sqrt{10}$.]The point of intersection of the two lines are ( $- 1, - 2$ )Distance $PM = \sqrt{10}$Hence the required line is one which passes through $( - 1, - 2)$ and is ⟂ to P.M. $\Rightarrow B$

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