If $a$ and $b$ are two arbitrary constants, then the straight line $(a-2b)x+(a+3b)y+3a+4b=0$ will pass through
A. $(-1,-2)$
B. $(1,2)$
C. $(-2,-3)$
D. $(2,3)$

Hint: In this question, we are to find the point through which the given straight-line passes. For that, we need to simplify the given equation and there we get a family of straight lines. So, the point of intersection of these lines will be the required point.

Formula Used: The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
The equation of a straight line passing through $(x,y)$ is $ax+by+c=0$.

Complete step by step solution: Given the equation of the straight line is
$(a-2b)x+(a+3b)y+3a+4b=0$
On simplifying,
$\begin{align}
  & \Rightarrow ax-2bx+ay+3by+3a+4b=0 \\
 & \Rightarrow a(x+y+3)+b(-2x+3y+4)=0 \\
\end{align}$
From the above, the straight lines we obtained are
$\begin{align}
  & x+y+3=0\text{ }...(1) \\
 & -2x+3y+4=0\text{ }...(2) \\
\end{align}$
The point of intersection of these two lines, is the required point through which the given line passes.
From (1),
$\begin{align}
  & x+y+3=0 \\
 & \Rightarrow x=-y-3\text{ }...(3) \\
\end{align}$
Substituting (3) in (2) we get,
$\begin{align}
  & -2x+3y+4=0 \\
 & \Rightarrow -2(-y-3)+3y+4=0 \\
 & \Rightarrow 2y+6+3y+4=0 \\
 & \Rightarrow 5y+10=0 \\
 & \Rightarrow y=-2 \\
\end{align}$
Then,
$\begin{align}
  & x=-y-3 \\
 & \Rightarrow x=-(-2)-3 \\
 & \Rightarrow x=-1 \\
\end{align}$
Therefore, the required point is $(-1,-2)$.

Option ‘A’ is correct

Note: Here we need to remember that the point of intersection of the straight lines obtained from the given equation is the required point through which the line passes.