Question

Find the equation of the perpendicular bisector of the line segment joining points $(7,1)$ and $(3,5)$

Answer 1

Step 1: Find the slope of the perpendicular bisector.

It is given that the line passes through the points $A(7,1)$ and $B(3,5)$

Let $m_1$is the slope of the line $AB$ and $m_2$is the slope of the perpendicular bisector $CD$

As $CD$ is a perpendicular bisector so the point $P$ will be the midpoint of the line $AB$

$\therefore$The coordinates of the midpoint of the line $AB$ $=\left(\frac{7+3}{2},\frac{5+1}{2}\right)=\left(5,3\right)$

We know that the slope of the line passes through the point $(x_1,y_1)$ and $\left(x_2,y_2\right)$is $m=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}$

$\therefore$The slope of the line $AB$ is $m_1$$=\frac{5-1}{3-7}=-1$

As we know when two lines are perpendicular, so $m_1\times m_2=-1$

$\Rightarrow m_2=\frac{-1}{m_1}=\frac{-1}{-1}=1$

Hence slope of a perpendicular bisector $CD=1$

Step 2: Find the equation of the perpendicular bisector.

We know that the equation of line passes through the point$\left(x_1,y_1\right)$ with slope $m$is $\left(y-y_1\right)=m\left(x-x_1\right)$

As the perpendicular bisector $CD$ passes through the point $\left(5,3\right)$

$$\begin{gathered} \Rightarrow (y–3)=1(x–5) \\ \Rightarrow y–3=x–5 \\ \Rightarrow x–y–2=0 \end{gathered}$$

Hence the required equation of the perpendicular bisector of the line segment joining points $(7,1)$ and $(3,5)$ is $x-y-2=0$