How do you find the midpoint of [ 11/3, 11/2] and [4/3, 3/2]?

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How do you find the midpoint of [ 11/3, 11/2] and [4/3, 3/2]?

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Answer 1 · 2016-03-12T16:07:48

$P_{m} = {\frac{55}{12}, \frac{17}{12}}$ $P_m={55/12,17/12}$

Explanation:

$P_{1} = [\frac{11}{3}, \frac{11}{2}] P_{2} = [\frac{4}{3}, \frac{3}{2}]$ $P_1=[11/3,11/2]" "P_2=[4/3,3/2]$
$P_{m} : Midpoint$ $P_m:" Midpoint"$
$P_{m x} = \frac{P_{1 x} + P_{2 x}}{2} = \frac{\frac{11}{3} + \frac{11}{2}}{2} = \frac{\frac{22 + 33}{6}}{2} = \frac{55}{12}$ $P_"m x"=(P_"1 x"+P_"2 x")/2=(11/3+11/2)/2=((22+33)/6)/2=55/12$
$P_{m y} = \frac{P_{1 y} + P_{2 y}}{2} = \frac{\frac{4}{3} + \frac{3}{2}}{2} = \frac{\frac{8 + 9}{6}}{2} = \frac{17}{12}$ $P_"m y"=(P_"1 y"+P_"2 y")/2=(4/3+3/2)/2=((8+9)/6)/2=17/12$
$P_{m} = {\frac{55}{12}, \frac{17}{12}}$ $P_m={55/12,17/12}$

Answer 2 · 2016-03-12T16:08:54

$(\frac{5}{2}, \frac{7}{2})$ $(5/2,7/2)$

Explanation:

To get the $x$ $x$ -coordinate of the midpoint, find the average of the $x$ $x$ -coordinates.
$\frac{\frac{11}{3} + \frac{4}{3}}{2} = \frac{\frac{15}{3}}{2} = \frac{5}{2}$ $(11/3+4/3)/2=(15/3)/2=5/2$
To get the $y$ $y$ -coordinate of the midpoint, find the average of the $y$ $y$ -coordinates.
$\frac{\frac{11}{2} + \frac{3}{2}}{2} = \frac{\frac{14}{2}}{2} = \frac{7}{2}$ $(11/2+3/2)/2=(14/2)/2=7/2$

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How do you find the midpoint of [ 11/3, 11/2] and [4/3, 3/2]?

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