Answer:
[tex] M(29, -12) [/tex]
Explanation:
Given:
[tex] H(4\frac{1}{2}, -3\frac{1}{4}), X(2\frac{3}{4}, -2\frac{3}{4}) [/tex]
Required:
Coordinates of the Midpoint of HX
Solution:
Midpoint (M) of HX, for [tex] H(4\frac{1}{2}, -3\frac{1}{4}), X(2\frac{3}{4}, -2\frac{3}{4}) [/tex] is given as
[tex] M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) [/tex].
Let [tex] H(4\frac{1}{2}, -3\frac{1}{4}) = (x_1, y_1) [/tex]
[tex] X(2\frac{3}{4}, -2\frac{3}{4}) = (x_2, y_2) [/tex]
Thus:
[tex] M(\frac{4\frac{1}{2} + 2\frac{3}{4}}{2}, \frac{-3\frac{1}{4} + (-2\frac{3}{4})}{2}) [/tex]
Convert to improper fractions
[tex] M(\frac{\frac{9}{2} + \frac{11}{4}}{2}, \frac{-\frac{13}{4} + (-\frac{11}{4})}{2}) [/tex]
[tex] M(\frac{\frac{18 + 11}{2}}{2}, \frac{\frac{-13 - 11}{4}}{2}) [/tex]
[tex] M(\frac{\frac{29}{2}}{2}, \frac{\frac{-24}{4}}{2}) [/tex]
[tex] M(\frac{29}{2}*2, \frac{-24}{4}*2 [/tex]
[tex] M(29, -12) [/tex]
