SOLUCIONES MATEMÁTICAS COESMAR: PUNTO MEDIO DE UN SEGMENTO DE RECTA

En la figura observe que los puntos (2 , 1) y $(-2\, ,\, \frac{1}{2})$ equidistan de los extremos de los segmentos a los cuales pertenecen $\overline{RS}$ y $\overline{PQ}$ , respectivamente.

Para el segmento horizontal $\overline{RS}$ , la abscisa del punto (2 , 1) es la semisuma de las abscisas de los extremos y la ordenada se mantiene. Para el segmento vertical $\overline{PQ}$ , la ordenada del punto $(-2\, ,\, \frac{1}{2})$ es la semisuma de las ordenadas de los extremos y la abscisa se mantiene.
Este mismo concepto puede aplicarse para otros segmentos de recta. Sean $P_{1}(x_{1},y_{1})$ las coordenadas de un extremo y $P_{2}(x_{2},y_{2})$ las coordenadas del otro extremo, tal como se muestra en la siguiente figura.

Con $P_{1}$ (-3, -2) y $P_{2}$ (5, 3) se verifica que:
$\frac{x_{1}+x_{2}}{2}=\frac{-3+5}{2}=1$
$\frac{y_{1}+y_{2}}{2}=\frac{-2+3}{2}=\frac{1}{2}$
Es decir, las coordenadas del punto M son (1, $\frac{1}{2}$ ), el cual equidista de $P_{1}$ y de $P_{2}$ . A continuación se enunciará y demostrará un teorema con el que se generaliza este resultado.

Teorema (Punto medio de un segmento de recta)

Si las coordenadas de los extremos del segmento $\overline{P_{1}P_{2}}$ son $P_{1}(x_{1},y_{1})$ y $P_{2}(x_{2},y_{2})$ , entonces las coordenadas del punto medio M de $\overline{P_{1}P_{2}}$ son:

$\left ( \frac{x_{1}+x_{2}}{2}\, ,\, \frac{y_{1}+y_{2}}{2} \right )$

Demostración:

En la ecuación de la recta L que contiene a los puntos $P_{1}(x_{1},y_{1})$ y $P_{2}(x_{2},y_{2})$ es:

$y-y_{1}=\left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x-x_{1})$

Debido a que el punto $M(x_{0},y_{0})$ pertenece a la recta L, tenemos:
$x_{0}-y_{1}=\left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x_{0}-x_{1})$
$\Rightarrow x_{0}=\left [\left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x_{0}-x_{1}) \right ]+y_{1}$ (1)
Como $M(x_{0},y_{0})$ es el punto medio entre $P_{1}$ y $P_{2}$ :
$d(P_{1},M)=d(M,P_{2})$

Aplicando la definición de distancia entre dos puntos, tenemos:
$\sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}=\sqrt{(x_{2}-x_{0})^{2}+(y_{2}-y_{0})^{2}}$

Elevando al cuadrado y transponiendo términos:

$(x_{1}-x_{0})^{2}-(x_{2}-x_{0})^{2}=(y_{2}-y_{0})^{2}+(y_{1}-y_{0})^{2}$

Factorizando:

$\left [(x_{1}-x_{0})+(x_{2}-x_{0}) \right ]\left [(x_{1}-x_{0})-(x_{2}-x_{0}) \right ]$
$=\left [(y_{2}-y_{0})+(y_{1}-y_{0}) \right ]\left [(y_{2}-y_{0})-(y_{1}-y_{0}) \right ]$

Simplificando:

$\left [ (x_{1}+x_{2})-2x_{0} \right ](x_{1}-x_{2})=\left [ (y_{1}+y_{2})-2y_{0} \right ](y_{2}-y_{1})$

Se reemplaza el valor de $y_{0}$ descrito por la ecuación (1), y se obtiene:

$\left [ (x_{1}+x_{2})-2x_{0} \right ](x_{1}-x_{2})$
$=\left [ (y_{1}+y_{2})-2\left (\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x_{0}-x_{1})-2y_{1} \right ](y_{2}-y_{1})$

A continuación se despeja $x_{0}$ :

$\left [2x_{0}- (x_{1}+x_{2}) \right ](x_{2}-x_{1})$
$=\left [ (y_{1}+y_{2})-2\left (\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x_{0}-x_{1})-2y_{1} \right ](y_{2}-y_{1})$

$\left [2x_{0}- (x_{1}+x_{2}) \right ](x_{2}-x_{1})^{2}$
$=\left [ (y_{1}+y_{2})({x_{2}-x_{1}})-2({y_{2}-y_{1}})(x_{0}-x_{1})-2y_{1}({x_{2}-x_{1}}) \right ]$
$(y_{2}-y_{1})$

$2x_{0}(x_{2}-x_{1})^{2}-(x_{2}-x_{1})(x_{2}-x_{1})^{2}$
$=(y_{2}+y_{1})(y_{2}-y_{1})(x_{2}-x_{1})-2(y_{2}-y_{1})^{2}(x_{0}+x_{1})$
$-2y_{1}(x_{2}+x_{1})(y_{2}+y_{1})$

$2x_{0}(x_{2}-x_{1})^{2}+2(y_{2}-y_{1})^{2}(x_{0}-x_{1})$
$=\left [ (y_{2}+y_{1})(y_{2}-y_{1})(x_{2}-x_{1})-2y_{1}(y_{2}-y_{1})(x_{2}-x_{1}) \right ]$
$+(x_{1}+x_{2})(x_{2}-x_{1})^{2}$

$2x_{0}(x_{2}-x_{1})^{2}+2x_{0}(y_{2}-y_{1})^{2}-2x_{1}(y_{2}-y_{1})^{2}$
$=\left [ (y_{2}-y_{1})(x_{2}-x_{1})(y_{2}+y_{1}-2y_{1}) \right ](x_{1}-x_{2})(x_{2}-x_{1})^{2}$

$2x_{0}(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$
$=(y_{2}-y_{1})(x_{2}-x_{1})(y_{2}-y_{1})+(x_{1}+x_{2})(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

$2x_{0}\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]$
$=(y_{2}-y_{1})^{2}(x_{2}-x_{1})+(x_{1}-x_{2})(x_{2}-x_{1})^{2}+2x_{1}(y_{2}-y_{1})^{2}$

$2x_{0}\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]$
$=\left [ (y_{2}-y_{1})^{2}(x_{2}-x_{1})+2x_{1}(y_{2}-y_{1})^{2} \right ]+(x_{1}+x_{2})(x_{2}-x_{1})^{2}$

$2x_{0}\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]$
$=\left [ (y_{2}-y_{1})^{2}(x_{2}-x_{1}+2x_{1}) \right ]+(x_{1}+x_{2})(x_{2}-x_{1})^{2}$

$2x_{0}\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]$
$=(y_{2}-y_{1})^{2}(x_{1}-x_{2})+(x_{1}+x_{2})(x_{2}-x_{1})^{2}$
$2x_{0}\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]=(x_{1}+x_{2})\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]$

Simplificando el factor $\left [(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} \right ]$ de ambos miembros, se tiene: $2x_{0}=x_{1}+x_{2}$ .

Luego, $x_{0}=\frac{x_{1}+x_{2}}{2}$

Para encontrar $y_{0}$ se reemplaza en (1):

$y_{0}=\left [ \left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right ) \left ( \frac{x_{1}+x_{2}}{2} -x_{1}\right )\right ]+y_{1}$
$y_{0}=\left [ \left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right ) \left ( \frac{x_{1}+x_{2}-2x_{1}}{2} \right )\right ]+y_{1}$
$y_{0}=\left [ \left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right ) \left ( \frac{x_{2}-x_{1}}{2} \right )\right ]+y_{1}$
$y_{0}=\frac{y_{2}-y_{1}}{2}+y_{1}$
$y_{0}=\frac{y_{2}-y_{1}+2y_{1}}{2}$
$y_{0}=\frac{y_{1}+y_{2}}{2}$ $\therefore$ $M=\left ( \frac{x_{1}+x_{2}}{2}\, ,\, \frac{y_{1}+y_{2}}{2} \right )$

Con esto se demuestra que el punto M equidista de los extremos del segmento $\overline{P_{1}P_{2}}$ .

Ejemplo.

Si M es el punto medio de $\overline{AB}$ , donde (-4 , -2) son las coordenadas de A y (2 , 1) son las coordenadas de M, determine las coordenadas de B.