# How do you find the midpoint of the line segment with endpoints (sqrt50,1) and (sqrt2,- 1)?

Dec 17, 2016

$\left(3 \sqrt{2} , 0\right)$

#### Explanation:

If the end points are $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$

then the midpoint is = $\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}$

Here $\left({x}_{1} , {y}_{1}\right) = \left(\sqrt{50} , 1\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right) = \left(\sqrt{2} , - 1\right)$

So midpoint = $\frac{\sqrt{50} + \sqrt{2}}{2} , \frac{1 + \left(- 1\right)}{2}$

or, $\frac{\sqrt{5 \cdot 5 \cdot 2} + \sqrt{2}}{2} , \frac{1 - 1}{2}$

or, $\frac{5 \sqrt{2} + \sqrt{2}}{2} , \frac{0}{2}$

or,$\frac{\sqrt{2} \left(5 + 1\right)}{2} , 0$

or, $\frac{6 \sqrt{2}}{2} , 0$

or, $3 \sqrt{2} , 0$

Dec 17, 2016

Mid point P_("mean") ->(x_("mean"),y_("mean"))=(3sqrt(2),0)

#### Explanation:

The mid point is the mean values

Let point 1 be ${P}_{1} = \left({x}_{1} , {y}_{1}\right) = \left(\sqrt{50} , 1\right)$
Let point 2 be P_2=(x_2,y_2)=(sqrt(2),-1))

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Consider ${x}_{1} = \sqrt{50}$

Note that $2 \times 25 = 50$ but $25 = {5}^{2}$ so we have:

${x}_{1} = \sqrt{2 \times {5}^{2}} = 5 \sqrt{2}$

Thus ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(5 \sqrt{2} , 1\right)$

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$\textcolor{b l u e}{\text{Determine the mean values}}$

${x}_{\text{mean}} \to \overline{x} = \frac{5 \sqrt{2} + \sqrt{2}}{2} = 3 \sqrt{2}$

${y}_{\text{mean}} \to \overline{y} = \frac{1 - 1}{2} = 0$
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Mid point P_("mean") ->(x_("mean"),y_("mean"))=(3sqrt(2),0)