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

Uriah Molina
2022-11-21
Answered

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

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AtticaPlotowvi

Answered 2022-11-22
Author has **18** answers

Step 1

Let mid point be $P}_{\text{mid}$

Let first point be ${P}_{1}\to ({x}_{1},{y}_{1})\to (-4,-3)$

Then $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=\frac{{P}_{2}+{P}_{1}}{2}$ ( mean value)

$\Rightarrow \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=(\frac{{x}_{2}+{x}_{1}}{2},\frac{{y}_{2}+{y}_{1}}{2})$

$\Rightarrow \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=(\frac{4+(-4)}{2},\frac{(-8)+(-3)}{2})=(0,-\frac{11}{2})$

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 $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{2}-{P}_{1}\to ({x}_{2}-{x}_{1})=(4-(-4))=+8$

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 ${P}_{1}+4=-4+4=0$

Let mid point be $P}_{\text{mid}$

Let first point be ${P}_{1}\to ({x}_{1},{y}_{1})\to (-4,-3)$

Then $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=\frac{{P}_{2}+{P}_{1}}{2}$ ( mean value)

$\Rightarrow \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=(\frac{{x}_{2}+{x}_{1}}{2},\frac{{y}_{2}+{y}_{1}}{2})$

$\Rightarrow \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{\text{mid}}=(\frac{4+(-4)}{2},\frac{(-8)+(-3)}{2})=(0,-\frac{11}{2})$

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 $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}{P}_{2}-{P}_{1}\to ({x}_{2}-{x}_{1})=(4-(-4))=+8$

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 ${P}_{1}+4=-4+4=0$

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