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2.4: Mifano iliyowekwa

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    164622

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    Katika sehemu hii, tumia fomu ya umbali\(d = \sqrt{(x_2 − x_1) ^2 + (y_2 − y_1) ^2}\) ili kupata urefu wa makundi ya mstari.

    Kumbuka: pointi tatu\(A\)\(B\), na\(C\) ni collinear, au kwa maneno mengine, pointi tatu ziko kwenye mstari huo, ikiwa jumla ya urefu wa makundi yoyote mawili ya mstari kuunganisha pointi, ni sawa na urefu wa sehemu iliyobaki ya mstari. Hiyo ni,\(AB + BC = AC\) au,\(AB + BC = AC\) au,\(AB + AC = BC\) au\(AC + BC = AB\).

    Mfano Template:index

    Kuamua kama pointi tatu zilizopewa ni collinear.

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

    Suluhisho

    Kwanza kupata makundi\(AB\),\(BC\), na\(AC\). Ili kufanya hivyo, tafuta umbali kati ya pointi\(A\) na\(B\),\(B\) na\(C\),\(A\) na\(C\).

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

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

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

    Hivyo,

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

    Tangu Hivyo,\(AB + BC = AC\) basi pointi tatu ni collinear.

    Zoezi Template:index
    1. Kuamua kama pointi zifuatazo ni collinear.
      1. \(A(4,-1)\quad B(5,-2) \quad C(1,2)\)
      2. \(A(2,-2)\quad B(3,1)\quad C(2,1)\)