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

Mar 3, 2016

${P}_{\text{mid}} \to \left(0 , - \frac{11}{2}\right)$

#### Explanation:

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

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

Let second point be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(4 , - 8\right)$

Then" "P_("mid") = (P_2+P_1)/2..... ( mean value)

=> " "P_("mid")=((x_2+x_1)/2 , (y_2+y_1)/2 )

=> " "P_("mid")= ((4+(-4))/2 ,((-8)+(-3))/2) = (0,-11/2)

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$\textcolor{b l u e}{\text{Something to think about}}$

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 $\text{ } {P}_{2} - {P}_{1} \to \left({x}_{2} - {x}_{1}\right) = \left(4 - \left(- 4\right)\right) = + 8$
Count from P_1" to "x_("mid") = +8/2 = 4

Actual $x$ value is ${P}_{1} + 4 = - 4 + 4 = 0$