Using the midpoint concept, it is found that:
- The midpoint of the line segment of endpoints [tex](2, \sqrt{3})[/tex] and [tex](-6, 5\sqrt{3})[/tex] is [tex](-2, 3\sqrt{3})[/tex].
- The midpoint of the line segment of endpoints [tex]\left(-\frac{5}{3}, \frac{2}{5}\right)[/tex] and [tex]\left(\frac{1}{3}, \frac{3}{2}\right)[/tex] is: [tex]\left(-\frac{2}{3}, \frac{19}{20}\right)[/tex]
The midpoint of a line segment is given by the mean of the coordinates of it's endpoints.
For the first segment, which endpoints [tex](2, \sqrt{3})[/tex] and [tex](-6, 5\sqrt{3})[/tex]:
[tex]x_M = \frac{2 - 6}{2} = -2[/tex]
[tex]y_M = \frac{\sqrt{3} + 5\sqrt{3}}{2} = 3\sqrt{2}[/tex]
The midpoint is [tex](-2, 3\sqrt{2})[/tex]
For the second segment, which endpoints [tex]\left(-\frac{5}{3}, \frac{2}{5}\right)[/tex] and [tex]\left(\frac{1}{3}, \frac{3}{2}\right)[/tex]:
[tex]x_M = \frac{-\frac{5}{3} + \frac{1}{3}}{2} = -\frac{4}{6} = -\frac{2}{3}[/tex]
[tex]y_M = \frac{\frac{2}{5} + \frac{3}{2}}{2} = \frac{\frac{4 + 15}{10}}{2} = \frac{19}{20}[/tex]
The midpoint is [tex]\left(-\frac{2}{3}, \frac{19}{20}\right)[/tex]
A similar problem is given at https://brainly.com/question/24352869
