Respuesta:
Calculemos las coordenadas del punto C de la medio del segmento AB:
\begin{gathered}x_{c} = \frac { xa + xb }{2} = \frac {-2 + 11 }{2} = \frac {9 }{2} = 4.5\\\\\\y_{c} =\frac { ya + yb }{2} = \frac{9 + (-3) }{2} =\frac { 6 }{2} = 3\end{gathered}
x
c
=
2
xa+xb
−2+11
9
=4.5
y
ya+yb
9+(−3)
6
=3
Coordenadas del punto medio (4.5, 3)
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Verified answer
Respuesta:
Calculemos las coordenadas del punto C de la medio del segmento AB:
\begin{gathered}x_{c} = \frac { xa + xb }{2} = \frac {-2 + 11 }{2} = \frac {9 }{2} = 4.5\\\\\\y_{c} =\frac { ya + yb }{2} = \frac{9 + (-3) }{2} =\frac { 6 }{2} = 3\end{gathered}
x
c
=
2
xa+xb
=
2
−2+11
=
2
9
=4.5
y
c
=
2
ya+yb
=
2
9+(−3)
=
2
6
=3
Coordenadas del punto medio (4.5, 3)