Question

determine a equação segmentária da reta que passa pelos pontos A(3,2) B(-1,4)

Answer 1

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$\large\begin{array}{l} \textsf{Considere dois pontos }\\\\A(x_{_A},\,y_{_A})\textsf{ e }B(x_{_B},\,y_{_B})\textsf{ do plano,}\\\\\textsf{com }x_{_A}\ne x_{_B}\textsf{ e }y_{_A}\ne y_{_B}.\\\\\\ \textsf{Considere tamb\'em que os pontos }A\textsf{ e }B\textsf{ n\~ao s\~ao}\\\textsf{a origem do sistema (o ponto }(0,\,0)\textsf{)}. \end{array}$


$\large\begin{array}{l} \textsf{Desejamos encontrar as coordenadas dos pontos }P\textsf{ e }Q,\\ \textsf{sendo }P\textsf{ um ponto sobre o eixo das abscissas (eixo x) e}\\Q\textsf{ um ponto sobre o eixo das ordenadas (eixo y)}\\\\ P(a,\,0)~\textsf{ e }~Q(0,\,b)\\\\ \textsf{de modo que }A,\,B,\,P\textsf{ e }Q\textsf{ estejam sobre uma mesma}\\\textsf{reta }r. \end{array}$


$\large\begin{array}{l} \textsf{Podemos usar dois dos pontos acima para calcular o}\\\textsf{coeficiente angular }m\textsf{ desta reta:}\\\\ m=\dfrac{y_{_B}-y_{_A}}{x_{_B}-x_{_A}}\qquad\quad\mathbf{(i)} \end{array}$


•   $\large\textsf{Usando os pontos }A\textsf{ e }P:$

$\large\begin{array}{l} m=\dfrac{y_{_P}-y_{_A}}{x_{_P}-x_{_A}}\\\\ m=\dfrac{0-y_{_A}}{a-x_{_A}}\\\\ m=\dfrac{-y_{_A}}{a-x_{_A}}\\\\ m(a-x_{_A})=-y_{_A} \end{array}$

$\large\begin{array}{l} a-x_{_A}=-\,\dfrac{y_{_A}}{m}\\\\ a=x_{_A}-\,\dfrac{y_{_A}}{m}\qquad\quad\mathbf{(ii)} \end{array}$


•   $\large\textsf{Usando os pontos }B\textsf{ e }Q:$

$\large\begin{array}{l} m=\dfrac{y_{_Q}-y_{_B}}{x_{_Q}-x_{_B}}\\\\ m=\dfrac{b-y_{_B}}{0-x_{_B}}\\\\ m=\dfrac{b-y_{_B}}{-x_{_B}} \end{array}$

$\large\begin{array}{l}-mx_{_B}=b-y_{_B}\\\\ b=y_{_B}-mx_{_B}\qquad\quad\mathbf{(iii)} \end{array}$


$\large\begin{array}{l} \textsf{A equa\c{c}\~ao segment\'aria da reta }r\textsf{ \'e dada por}\\\\ \boxed{\begin{array}{c}r:~~\dfrac{x}{a}+\dfrac{y}{b}=1 \end{array}}\qquad\quad(a\ne 0\textsf{ e }b\ne 0) \end{array}$

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$\large\begin{array}{l} \textsf{Para esta tarefa, temos}\\\\ A(3,\,2)\textsf{ e }B(-1,\,4). \end{array}$


•   $\large\textsf{Calculando o coeficiente angular:}$

$\large\begin{array}{l} m=\dfrac{y_{_B}-y_{_A}}{x_{_B}-x_{_A}}\\\\ m=\dfrac{4-2}{-1-3}\\\\ m=\dfrac{2}{-4}\\\\ m=-\,\dfrac{1}{2}\qquad\quad\checkmark\\\\ \end{array}$


•   $\large\textsf{Encontrando o ponto de interse\c{c}\~ao da reta com}$
$\large\textsf{o eixo x:}$

$\large\begin{array}{l} a=x_{_A}-\dfrac{y_{_A}}{m}\\\\ a=3-\dfrac{2}{-\,\frac{1}{2}}\\\\ a=3-2\cdot (-2)\\\\ a=3+4\\\\ a=7\qquad\quad\checkmark \end{array}$


•   $\large\textsf{Encontrando o ponto de interse\c{c}\~ao da reta com}$
$\large\textsf{o eixo y:}$

$\large\begin{array}{l} b=y_{_B}-mx_{_B}\\\\ b=4-\left(-\dfrac{1}{2}\right)\cdot (-1)\\\\ b=4-\dfrac{1}{2} \end{array}$

$\large\begin{array}{l} b=\dfrac{8}{2}-\dfrac{1}{2}\\\\ b=\dfrac{8-1}{2}\\\\ b=\dfrac{7}{2}\qquad\quad\checkmark \end{array}$


•   $\large\textsf{Equa\c{c}\~ao segment\'aria da reta procurada:}$

$\large\begin{array}{l} r:~~\dfrac{x}{a}+\dfrac{y}{b}=1\\\\ \boxed{\begin{array}{c}r:~~\dfrac{x}{7}+\dfrac{\;y\;}{\frac{7}{2}}=1 \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: equação segmentária reta ponto coeficiente angular interseção eixo geometria analítica