How many perpendicular bisectors can be constructed for a line?

Hint: To solve this question, we need to consider the general equation of a line. Thus, the characteristics of a line are a point which lies on it, and its slope. A line is uniquely determined by a point and its slope.

Complete step-by-step solution:
We know that the perpendicular bisector for a given line is a line bisecting the given line and also which is perpendicular to it. This means that the perpendicular bisector passes through the midpoint of the given line and is perpendicular to it.
Let the midpoint of the given line be $\left( {h,k} \right)$, and its slope be equal to $m$. By definition the perpendicular bisector for this line must be perpendicular to it. We know that the product of the slopes of two perpendicular lines is equal to $ - 1$. Therefore, the slope of the perpendicular bisector of the given line is given by
${m_p}m = - 1$
$ \Rightarrow {m_p} = - \dfrac{1}{m}$
Also, since the midpoint of the given line is assumed at the point $\left( {h,k} \right)$, the perpendicular bisector must pass through it. Now, the general equation of a line is given by
$y - {y_1} = m\left( {x - {x_1}} \right)$
Substituting ${x_1} = h$, ${y_1} = k$, and $m = - \dfrac{1}{m}$, we get
$y - k = - \dfrac{1}{m}\left( {x - h} \right)$
The above equation completely determines the equation of the perpendicular bisector, and does not involve any independent constant.

Hence, we can conclude that there is only one perpendicular bisector which can be constructed for a given line.

Note:
We can find the number of perpendicular bisectors without using the general equation of a line. The term “perpendicular” indicates that the perpendicular bisector line must be at 90 degress to the given line. So it includes all the family of lines perpendicular to the given line. While the term “bisector” selects one member of this family of perpendicular lines, which passes through its midpoint. By drawing a perpendicular bisector, we can easily divide the sine into two equal halves.