# How do you find the midpoint of (3,1) and (7,-5)?

Mar 3, 2016

${P}_{\text{mid}} = \left(2 , - 3\right)$

#### Explanation:

Let the first point be$\text{ } {P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(3 , 1\right)$

Let the second point be$\text{ } {P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(7 , - 5\right)$

Let the mid point be " "P_("mid")

The mid point between ${P}_{1} \text{ and } {P}_{2}$ is such that its coordinates are the mid point of the $x ' s$ and the mid point of the $y ' s$

${P}_{\text{mid}} \to \frac{{P}_{2} - {P}_{1}}{2} \to \left(x , y\right) \to \left(\frac{{x}_{2} - {x}_{1}}{2} , \frac{{y}_{2} - {y}_{1}}{2}\right)$

$\text{ } = \left(\frac{7 - 3}{2} , \frac{\left(- 5\right) - 1}{2}\right)$

${P}_{\text{mid}} = \left(2 , - 3\right)$