Given the family of lines $a(2x+y+4)+b(x-2y-3)=0$. The numbers of lines belonging to the family at a distance $\sqrt { 10 }$ from any point $(2,-3)$ is

soln : Given,

$\begin{array}{l} 2 x+y+4=0\ {line 1}\\ x-2 y-3=0 \quad \text { Line 2 } \end{array}$

NOM,

$\begin{array}{l} 4 x+2 y+8=0 \quad[\because \text { multiply by } 2] \\ x-2 y-3=0 \\ 5 x+5=0 \Rightarrow x =-1 \end{array}$

put value of $x$ an Line 2

$\begin{array}{c} (-1)-2 y-3=0 \\ y=-2 \end{array}$

So, Lines passes through intersecting points $[-1,-2]$

Now,

$\begin{array}{l} \left(y-y_{1}\right)=m\left(x-x_{1}\right) \\ y+2=m(x+1) \\ m x-y+m-2=0 \end{array}$

Distance between $[2,-3]$ and $m x-y+m-2 = 0$ is $\sqrt{10}$. [ $\because$ Given]

$\begin{aligned} \text { Distance } & =\mid \frac{A m+B n+C \mid}{\sqrt{A^{2}+B^{2}}} \\ \sqrt{10} & =\left|\frac{2 m+3+m-2}{\sqrt{m^{2}+1}}\right| \\ \sqrt{10} & =\left|\frac{3 m+1}{\sqrt{m^{2}+1}}\right| \\ 10\left(m^{2}+1\right) & =(3 m+1)^{2} \quad[\because \text { sq both side] } \\ 10 m^{2}+10 & =9 m^{2}+1+6 m \\ m^{2}-6 m+9=0 & =0 \\ (m-3)^{2} & =0 \\ m & =3 . \end{aligned}$

$\therefore M$ has only one value i.e $m = 3$

so, only one line is possible. the correct answer is B