# How do you find the midpoint of the line segment joining (2,-3) and (8, -1)?

Mar 28, 2017

See the entire solution process below:

#### Explanation:

The formula to find the mid-point of a line segment give the two end points is:

$M = \left(\frac{\textcolor{red}{{x}_{1}} + \textcolor{b l u e}{{x}_{2}}}{2} , \frac{\textcolor{red}{{y}_{1}} + \textcolor{b l u e}{{y}_{2}}}{2}\right)$

Where $M$ is the midpoint and the given points are:

$\left(\textcolor{red}{\left({x}_{1} , {y}_{1}\right)}\right)$ and $\left(\textcolor{b l u e}{\left({x}_{2} , {y}_{2}\right)}\right)$

Substituting the values from the points in the problem gives:

$M = \left(\frac{\textcolor{red}{2} + \textcolor{b l u e}{8}}{2} , \frac{\textcolor{red}{- 3} + \textcolor{b l u e}{- 1}}{2}\right)$

$M = \left(\frac{10}{2} , - \frac{4}{2}\right)$

$M = \left(5 , - 2\right)$

Mar 28, 2017

$\left(5 , - 2\right)$

#### Explanation:

The mid=point of a line segment is the arithmetic mean of the co-ordinates

$\text{midpoint of } \left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$

$\text{is given by } M \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

$\text{ for "(2,-3),(8,-1) " we have }$

$\left(\frac{2 + 8}{2} , \frac{- 3 + - 1}{2}\right)$

$\left(\frac{10}{2} , - \frac{4}{2}\right) = \left(5 , - 2\right)$