Question

Find the equation of the perpendicular bisector of the line joining the points $\left(1,3\right)$ and $\left(3,1\right)$.

Answer 1

Step 1: Finding the slope of the line joining the points $\left(1,3\right)$ and $\left(3,1\right)$:
Slope of a line is given by:

$m=\frac{y_2-y_1}{x_2-x_1}$
$\Rightarrow m_1=\frac{1-3}{3-1}$ (where $m_1$ is the slope of the line joining the points $\left(1,3\right)$ and $\left(3,1\right)$)
$\Rightarrow m_1=-1$

Step 2: Finding the slope of the perpendicular bisector:
Given that the required line is the perpendicular bisector of the line joining the points $\left(1,3\right)$ and $\left(3,1\right)$.

$\Rightarrow m_1\times m_2=-1$ (where $m_2$ is the slope of the required line)

$\Rightarrow -m_2=-1$

$\Rightarrow m_2=1$

Step 3: Finding the midpoint of $\left(1,3\right)$ and $\left(3,1\right)$:
Let $A\left(x_{},y_{}\right)$be the midpoint of $\left(1,3\right)$ and $\left(3,1\right)$.

using mid point formula $\left(x_{},y_{}\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$\Rightarrow \left(x_{},y_{}\right)=\left(\frac{1+3}{2},\frac{3+1}{2}\right)$

$\Rightarrow A\left(x_{},y_{}\right)=\left(2,2\right)$

Step 4: Finding the equation of the required line:
We know that, slope $m_2$ of the required line is $1$.
Also, the required line passes through the point $\left(2,2\right)$
Thus, equation of a line in slope-point form is given as:
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y-2=1\left(x-2\right)$
$\Rightarrow x-y=0$

Hence, equation of the required line is $x-y=0$