Question

Encontre a equação da mediatriz do segmento PQ , sendo P(1,2) e Q(-3,4) . Em seguida , escolha um ponto qualquer dessa mediatriz e mostre que ele equidista de P e Q. ???????

Answer 1

Temos os seguintes pontos

$P(1,\,2)~\text{ e }~Q(-3,\,4).$


$\bullet~~$ Encontrar o coeficiente angular $m$ do segmento $PQ:$

$m=\dfrac{\Delta y}{\Delta x}=\dfrac{y_{_{Q}}-y_{_{P}}}{x_{_{Q}}-x_{_{P}}}\\\\\\ m=\dfrac{4-2}{-3-1}\\\\\\ m=\dfrac{2}{-4}\\\\\\ \boxed{\begin{array}{c} m=-\,\dfrac{1}{2} \end{array}}$


$\bullet~~$ Encontrar o ponto médio $M(x_{_{M}},\,y_{_{M}})$ do segmento $PQ:$

$x_{_{M}}=\dfrac{x_{_{P}}+x_{_{Q}}}{2}\\\\\\ x_{_{M}}=\dfrac{1+(-3)}{2}\\\\\\ x_{_{M}}=\dfrac{1-3}{2}\\\\\\ x_{_{M}}=\dfrac{-2}{2}\\\\\\ \boxed{\begin{array}{c}x_{_{M}}=-1 \end{array}}\\\\\\\\ y_{_{M}}=\dfrac{y_{_{P}}+y_{_{Q}}}{2}\\\\\\ y_{_{M}}=\dfrac{2+4}{2}\\\\\\ y_{_{M}}=\dfrac{6}{2}\\\\\\ \boxed{\begin{array}{c}y_{_{M}}=3 \end{array}}$


O ponto médio do segmento $PQ$ é o ponto $M(-1,\,3).$


$\bullet~~$ Encontrar o coeficiente angular $m_{\perp}$ da reta mediatriz do segmento $PQ.$

(A reta mediatriz do segmento $PQ$ é a reta perpendicular ao segmento $PQ$ que passa pelo ponto médio deste segmento)


Como a reta mediatriz é perpendicular ao segmento $PQ,$ devemos ter

$m_{\perp}\cdot m=-1\\\\ m_{\perp}=-\,\dfrac{1}{m}\\\\\\ m_{\perp}=-\,\dfrac{1}{(-\frac{1}{2})}\\\\\\ m_{\perp}=-1\cdot (-2)\\\\ \boxed{\begin{array}{c} m_{\perp}=2 \end{array}}$


$\bullet~~$ Encontrando a equação da reta mediatriz do segmento $PQ:$

(A reta mediatriz $r$ é a reta que passa pelo ponto $M(-1,\,3)$ com coeficiente angular $m_{\perp}=2$ )

$r:~~y-y_{_{M}}=m_{\perp}\cdot (x-x_{_{M}})\\\\ r:~~y-3=2\cdot (x-(-1))\\\\ r:~~y-3=2\cdot (x+1)\\\\ r:~~y-3=2x+2\\\\ r:~~y=2x+2+3\\\\ \boxed{\begin{array}{c} r:~~y=2x+5 \end{array}}$


$\bullet~~$ Mostrar que um ponto $A(x_{_{A}},\,y_{_{A}})$ qualquer da reta mediatriz é equidistante de $P$ e $Q:$

Se $A$ pertence à reta mediatriz, então as coordenadas de $A$ devem satisfazer a equação da reta $r:$

$y_{_{A}}=2x_{_{A}}+5$


Calculando a distância de $A$ até $P:$

$d_{_{AP}}=\sqrt{(x_{_{P}}-x_{_{A}})^2+(y_{_{P}}-y_{_{A}})^2}\\\\ d_{_{AP}}=\sqrt{(1-x_{_{A}})^2+(2-y_{_{A}})^2}\\\\ d_{_{AP}}=\sqrt{(1-x_{_{A}})^2+(2-(2x_{_{A}}+5))^2}\\\\ d_{_{AP}}=\sqrt{(1-x_{_{A}})^2+(2-2x_{_{A}}-5)^2}\\\\ d_{_{AP}}=\sqrt{(1-x_{_{A}})^2+(-2x_{_{A}}-3)^2}\\\\ d_{_{AP}}=\sqrt{(x_{_{A}}-1)^2+(2x_{_{A}}+3)^2}\\\\ d_{_{AP}}=\sqrt{x_{_{A}}^2-2x_{_{A}}+1+4x_{_{A}}^2+12x_{_{A}}+9}\\\\ d_{_{AP}}=\sqrt{5x_{_{A}}^2+10x_{_{A}}+10}\\\\ d_{_{AP}}=\sqrt{5\,(x_{_{A}}^2+2x_{_{A}}+2)}~~~~~~\mathbf{(i)}$


Calculando a distância de $A$ até $Q:$

$d_{_{AQ}}=\sqrt{(x_{_{Q}}-x_{_{A}})^2+(y_{_{Q}}-y_{_{A}})^2}\\\\ d_{_{AQ}}=\sqrt{(-3-x_{_{A}})^2+(4-y_{_{A}})^2}\\\\ d_{_{AQ}}=\sqrt{(-3-x_{_{A}})^2+(4-(2x_{_{A}}+5))^2}\\\\ d_{_{AQ}}=\sqrt{(-3-x_{_{A}})^2+(4-2x_{_{A}}-5)^2}\\\\ d_{_{AQ}}=\sqrt{(-3-x_{_{A}})^2+(-2x_{_{A}}-1)^2}\\\\ d_{_{AQ}}=\sqrt{(x_{_{A}}+3)^2+(2x_{_{A}}+1)^2}\\\\ d_{_{AQ}}=\sqrt{x_{_{A}}^2+6x_{_{A}}+9+4x_{_{A}}^2+4x_{_{A}}+1}\\\\ d_{_{AQ}}=\sqrt{5x_{_{A}}^2+10x_{_{A}}+10}\\\\ d_{_{AQ}}=\sqrt{5\,(x_{_{A}}^2+2x_{_{A}}+2)}~~~~~~\mathbf{(ii)}$


Comparando $\mathbf{(i)}$ e $\mathbf{(ii)},$ concluímos que

$\boxed{\begin{array}{c}d_{_{AP}}=d_{_{AQ}} \end{array}}$


Assim, verificamos que um ponto qualquer da reta mediatriz é equidistante de $P$ e $Q.$