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

Mar 12, 2016

${P}_{m} = \left\{\frac{55}{12} , \frac{17}{12}\right\}$

Explanation:

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

Mar 12, 2016

$\left(\frac{5}{2} , \frac{7}{2}\right)$

Explanation:

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