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2.4: Exemplos aplicados

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    170184

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    Nesta seção, aplique a fórmula da distância\(d = \sqrt{(x_2 − x_1) ^2 + (y_2 − y_1) ^2}\) para encontrar os comprimentos dos segmentos de linha.

    Nota: Três pontos\(A\),\(B\), e\(C\) são colineares, ou em outras palavras, os três pontos estão na mesma linha, se a soma dos comprimentos de quaisquer dois segmentos de linha conectando os pontos for igual ao comprimento do segmento de linha restante. Ou seja,\(AB + BC = AC\) ou,\(AB + BC = AC\) ou,\(AB + AC = BC\) ou\(AC + BC = AB\).

    Exemplo Template:index

    Determine se os três pontos fornecidos são colineares.

    \(A(10, −4)\quad B(8, −2) \quad C(2, 4)\)

    Solução

    Primeiro\(AB\), encontre os segmentos\(BC\),\(AC\) e. Para fazer isso, encontre a distância entre os pontos\(A\)\(B\) e\(B\)\(C\),\(A\) e\(C\) e.

    \(\begin{aligned} \text{Segment AB }&=\text{ The distance between point A and Point B } \\ &= \sqrt{(8 − 10)^2 + [−2 − (−4)]^2} \\ &= \sqrt{(−2)^2 + (2)^2} \\&= \sqrt{ 8}\\&= 2\sqrt{2} \end{aligned}\)

    \(\begin{aligned} \text{Segment BC }&=\text{ The distance between point B and Point C } \\ &= \sqrt{(2 − 8)^2 + [4 − (−2)]^2 }\\ &= \sqrt{(−6)^2 + (6)^2} \\&= \sqrt{ 72 }\\&= 6\sqrt{ 2}\end{aligned}\)

    \(\begin{aligned} \text{Segment AC }&=\text{ The distance between point A and Point C }\\&= \sqrt{(2 − 10)^2 + [4 − (−4)]^2} \\&= \sqrt{(−8)^2 + (8)^2 }\\&= \sqrt{ 128 }\\&= 8\sqrt{ 2}\end{aligned}\)

    Assim,

    \(\begin{aligned} AB + BC &= 2\sqrt{ 2} + 6\sqrt{ 2 }\\&= 8\sqrt{ 2 } \\&= AC \end{aligned}\)

    Desde\(AB + BC = AC\) então, três pontos são colineares.

    Exercício Template:index
    1. Determine se os seguintes pontos são colineares.
      1. \(A(4,-1)\quad B(5,-2) \quad C(1,2)\)
      2. \(A(2,-2)\quad B(3,1)\quad C(2,1)\)