points are $A(2 a, 4)$ and $B(-2,3 b)$

Let $C(1,2 a+1)$ be the mid-point of $A B$.

$\mathrm{x}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \mathrm{y}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}$

$\Rightarrow 1=\frac{2 \mathrm{a}(-2)}{2}, 2 \mathrm{a}+1=\frac{4+3 \mathrm{~b}}{2}$

$\Rightarrow 2=2 a-2,4 a+2=4+3 b$

$\Rightarrow 2 a=2+2,4 a-3 b=4-2$

$$
\begin{aligned}
&\Rightarrow a=\frac{4}{2}, 4 a-3 b=2 \\
&\Rightarrow a=2,4 a-3 b=2
\end{aligned}
$$

Putting the value of a in the equation $4 a+3 b=2$,

$4(2)-3 b=2$

$\Rightarrow-3 b=2-8=-6$

$\Rightarrow \mathrm{b}=\frac{6}{3}=2$

Hence, $a=2$ and $b=2$