# How do you find the midpoint of (-4, -3),(4, -8)?

How do you find the midpoint of (-4, -3),(4, -8)?
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Step 1
Let mid point be ${P}_{\text{mid}}$
Let first point be ${P}_{1}\to \left({x}_{1},{y}_{1}\right)\to \left(-4,-3\right)$
Then $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=\frac{{P}_{2}+{P}_{1}}{2}$ ( mean value)
$⇒\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=\left(\frac{{x}_{2}+{x}_{1}}{2},\frac{{y}_{2}+{y}_{1}}{2}\right)$
$⇒\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=\left(\frac{4+\left(-4\right)}{2},\frac{\left(-8\right)+\left(-3\right)}{2}\right)=\left(0,-\frac{11}{2}\right)$
Step 2
Using the mean value is more straightforward than using
${P}_{1}+\frac{{P}_{2}-{P}_{1}}{2}$ which is the alternative.
Permit me to demonstrate just for the mid x value
Distance between is $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{2}-{P}_{1}\to \left({x}_{2}-{x}_{1}\right)=\left(4-\left(-4\right)\right)=+8$
Count from ${P}_{1}\phantom{\rule{1ex}{0ex}}\text{to}\phantom{\rule{1ex}{0ex}}{x}_{\text{mid}}=+\frac{8}{2}=4$
Actual x value is ${P}_{1}+4=-4+4=0$