Question

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

Answer 1

Step $1$: Simplification given data
Let the given points be $A(3,4)$ & $B(-1,2)$
Let $CD$ be the right bisector of line $AB$
Slope of a line joining points $(x_1,y_1)$ & $(x_2,y_2)$ is $=\frac{y_2-y_1}{x_2-x_1}$
Slope of $AB$ joining $(3,4)$ & $(-1,2)$ is
$\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}$
So, slope of $AB=\frac{1}{2}$
Since
$CD$ is the right bisector of line $AB$
$\therefore$ Line $CD\perp$ line $AB$
And, we know that if two lines are perpendicular, then their product of slope is $-1$
So, slope of $CD\times$ Slope of $AB=-1$
Slope of $CD=\frac{-1}{\text{Slope of}AB}=\frac{-1}{\frac{1}{2}}=-2$

Step $2$: Required equation of line
Point $P$ is the mid-point of line $AB$
We know that co-ordinates of mid-points of two points $(x_1,y_1)$ & $(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
So, co-ordinates of point $P=(\frac{-1+3}{2},\frac{2+4}{2})=(\frac{2}{2},\frac{6}{2})=(1,3)$
We know that equation of a line passes through $(x_1,y_1)$ & having slope of $m$ is $(y-y_1)=m(x-x_1)$
Equation of the $CD$ passing through point $P(1,3)$ & slope of $-2$ is
$\Rightarrow (y-3)=-2(x-1)$
$\Rightarrow y-3=-2x+2$
$\Rightarrow y+2x=2+3$
$\Rightarrow 2x+y=5$

Final answer:
Therefore, the required equation is $2x+y-5=0$.