Question

Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

Answer 1

Hint: We know that the right bisector of a line segment is passing through its midpoint and perpendicular to it. We will find the midpoint using the formula as follows:
If we have end points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\].
X coordinate of midpoint \[=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\]
Y coordinate of midpoint \[=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]
Also we will use the property that the product of the slope of a line and a line perpendicular to it is equal to minus one.

Complete step-by-step answer:
We have been asked to find the right bisector of the line segment the points (3, 4) and (-1, 2).
We know that the right bisector of the line segment is perpendicular and passing through the midpoint of the line segment.
We know the midpoint of line segment joining points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by:
X coordinate of midpoint \[=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\]
Y coordinate of midpoint \[=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\]
So midpoint of the line segment joining the points (3, 4) and (-1, 2) is given by:
X coordinate of midpoint \[=\dfrac{3-1}{2}=\dfrac{2}{2}=1\]
Y coordinate of midpoint \[=\dfrac{4+2}{2}=\dfrac{6}{2}=3\]
Hence the coordinate of the point is (1,3).
Also, we know that if we have two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] then slope is given by:
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
So the slope of line segment joining (3, 4) and (-1, 2) is given by:
\[m=\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}\]
Since we know that the product of a line and its perpendicular line is equal to minus one
Let the slope of right bisector be \[{{m}_{1}}\]
\[\begin{align}
  & \Rightarrow m{{m}_{1}}=-1 \\
 & \Rightarrow \dfrac{1}{2}\times {{m}_{1}}=-1 \\
 & \Rightarrow {{m}_{1}}=-2 \\
\end{align}\]
We know the equation of a line in slope and point form is given by:
\[y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\] where \[\left( {{x}_{1}},{{y}_{1}} \right)\] is a point through which the line passes and m is the slope.
So the equation of right bisector passing through (1,3) and having slope -2 is given by:
\[\begin{align}
  & y-3=-2\left( x-1 \right) \\
 & y-3=-2x+2 \\
 & y+2x-3-2=0 \\
 & y+2x-5=0 \\
\end{align}\]
Therefore, the required equation of the right bisector of the line segment is \[y+2x-5=0\].

Note: Be careful while finding the values of midpoint and the slope of the line segment as there is a chance of sign mistake during calculation. It must be remembered that while finding the midpoint, we add the coordinates and not subtract them. This is probably a common silly mistake that can be made. Also, remember the property that if two lines are perpendicular then the product of their slope is equal to minus one.