It is likely more helpful to set out the problem using vector forms for the relevant lines. The vector $ \ \vec{c} \ = \ \vec{AB} \ $ is $ \ \vec{c} \ = \ \vec{b} \ - \ \vec{a} \ $ (since $ \ \vec{a} \ + \ \vec{c} \ = \ \vec{b} \ $ ) . It appears that you are working in $ \ \mathbb{R}^2 \ $ , so we have
$$ \vec{a} \ = \ \langle x_1 \ , \ y_1 \rangle \ \ , \ \ \vec{b} \ = \ \langle x_2 \ , \ y_2 \rangle \ \ , \ \ \vec{c} \ = \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
It is convenient to use $ \ \vec{a} \ $ as the "initial" vector with a parameterization that places $ \ t = 0 \ $ there (point A) and places $ \ t = 1 \ $ at $ \ \vec{b} \ $ (point B) , with the direction given by $ \ \vec{c} \ $ . The vector $ \ \vec{c} \ $ then lies along the line AB having parametric equation
$$ \ \vec{r} \ = \ \vec{a} \ + \ t \ \vec{c} \ = \ \langle x_1 \ , \ y_1 \rangle \ + \ t \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
The midpoint is then located at $ \ t = \frac{1}{2} \ $ , said point being at the head of position vector $ \ \langle \frac{x_1 \ + \ x_2}{2} \ , $ $ \ \frac{y_1 \ + \ y_2}{2} \rangle \ $ .
The perpendicular bisector of $ \ \overline{AB} \ $ can be defined by a vector which has a zero "dot product" (scalar product) with $ \ \vec{c} \ $ . If we say that this line has a slope $ \ m \ $ , we can write its direction as $ \ \langle 1 \ , \ m \rangle \ $ , and so
$$ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \cdot \ \langle 1 \ , \ m \rangle \ \ = \ \ (x_2 - x_1 ) \ + \ m \ (y_2 - y_1) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ m \ \ = \ \ - \frac{x_2 - x_1}{y_2 - y_1} \ \ . $$
(Note that this is just what we expect from the "product of slopes of perpendicular lines" theorem, since the direction $ \ \vec{c} \ $ has slope $ \ \frac{y_2 - y_1}{x_2 - x_1} \ $ . )
Since $ \ y_2 - y_1 \ \neq \ 0 \ $ , we can then also write the direction for the perpendicular bisector as $ \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ $ . (It is also acceptable to attach the "minus sign" to the first component instead -- thus, $ \ \langle y_1 - y_2 \ , \ x_2 - x_1 \rangle \ $ -- which is a vector pointing in the opposite direction on the perpendicular bisector line.
Finally, we can establish a parametrization of the perpendicular bisector with the midpoint of $ \ \overline{AB} \ $ at $ \ \tau = 0 \ $ , making it the "initial" vector for the bisector. The vector form for this line may then be written as
$$ \vec{R} \ = \ \langle \frac{x_1 \ + \ x_2}{2} \ , \ \frac{y_1 \ + \ y_2}{2} \rangle \ + \ \tau \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ \ . $$
This method can be extended to $ \ \mathbb{R}^3 \ $ , except that the perpendicular bisector is ambiguous to the extent that it lies in a plane for which $ \ \vec{AB} \ $ is its normal. To pin things down, we would need to specify a point in that plane that the bisector passes through. (This is sometimes given as a textbook or exam problem.)
The equation
$$\vec r = \vec r_0 + t \vec v$$
is a parametric equation for a line. You can see $\vec r$ as a function of $t$:
$$\vec r(t) = r_0 + tv$$
and for any specific $t_0$, $\vec r(t_0)$ will be a vector pointing to a point on the line.
The whole line is the set of all points which are produced by $\vec r(t)$, i.e. the set
$$\{\vec r(t) \mid t \in \mathbb R\}.$$
Thus, it does not make sense to ask "how to solve for $t$?" as your question is phrased.
However, if you have a point $\vec p$ and ask the question "for which $t$ is $\vec r(t) = \vec p$?", that does make sense. In this case you solve the equation
$$\vec p = \vec r(t) = \vec r_0 + t \vec v$$
and this will be solvable only when $\vec p$ is on the line.
Best Answer
Another way to look at the situation is as follows.
We can make Cartesian coordinates for a three-dimensional Euclidean space by choosing an $x$ axis, a $y$ axis, and a $z$ axis, which define a coordinate system on this space. The point where the three axes intersect is the origin of this coordinate system and is given the name $(0,0,0)$ in the coordinate system. Each point $p$ in space, which is defined by its position in the space, has some coordinates $(p_x,p_y,p_z)$ in the coordinate system.
We can also define a three-dimensional vector space in which each vector $v$ is described by three coordinates $(v_x,v_y,v_z).$ A vector in this vector space doesn't really have a "position" in the sense that we think of the point $p$ in the Euclidean space as having a position, but the vector does have a length and a direction.
Because either a point or a vector can be described by three coordinates, it is convenient to write equations involving objects of three coordinates that are alternatively interpreted either as coordinates of points or coordinates of vectors. Hence in $$ \vec{r}(t) = \vec{a} + t\vec{v}, $$ we start with the three coordinates of a point $a$ through which the desired line passes and treat them as coordinates of a vector $\vec a.$ We take the three coordinates of a vector $\vec v$ parallel to the desired line, and add some multiple of this vector to $\vec a.$ The result is a vector $\vec{r}(t),$ whose three coordinates we interpret as the coordinates of a point $r(t)$ on the line.
In order to make sense out of this, we might use the concept of position vectors. The position vector of a point $p$ is the vector $\vec p$ equal to the distance and direction from the origin of coordinates to $p.$ Constructed this way, the vector $\vec p$ always has the same coordinates in the vector space as the point $p$ has in the Euclidean space, so it's easy to swap a vector for a point or a point for a vector in calculations.
We may even stop representing points as separate objects and talk only about their position vectors. This is often what is meant by the equation $\vec{r}(t) = \vec{a} + t\vec{v},$ where $\vec{r}(t)$ is a position vector of a point on the line and that is simply how we identify that point.