2.4: 应用示例
- Page ID
- 171177
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)在本节中，应用距离公式\(d = \sqrt{(x_2 − x_1) ^2 + (y_2 − y_1) ^2}\)来计算线段的长度。
注意：三个点\(A\)\(B\)、和\(C\)是共线的，换句话说，如果连接两点的任意两条线段的长度之和等于剩余线段的长度，则这三个点位于同一条线上。 即、\(AB + BC = AC\)\(AB + BC = AC\)或、\(AB + AC = BC\)或\(AC + BC = AB\)。
确定给定的三个点是否共线。
\(A(10, −4)\quad B(8, −2) \quad C(2, 4)\)
解决方案
首先找到分段\(AB\)\(BC\)、和\(AC\)。 为此，请找出点与\(B\)、\(A\)\(B\)\(A\)和\(C\)、之间的距离\(C\)。
\(\begin{aligned} \text{Segment AB }&=\text{ The distance between point A and Point B } \\ &= \sqrt{(8 − 10)^2 + [−2 − (−4)]^2} \\ &= \sqrt{(−2)^2 + (2)^2} \\&= \sqrt{ 8}\\&= 2\sqrt{2} \end{aligned}\)
\(\begin{aligned} \text{Segment BC }&=\text{ The distance between point B and Point C } \\ &= \sqrt{(2 − 8)^2 + [4 − (−2)]^2 }\\ &= \sqrt{(−6)^2 + (6)^2} \\&= \sqrt{ 72 }\\&= 6\sqrt{ 2}\end{aligned}\)
\(\begin{aligned} \text{Segment AC }&=\text{ The distance between point A and Point C }\\&= \sqrt{(2 − 10)^2 + [4 − (−4)]^2} \\&= \sqrt{(−8)^2 + (8)^2 }\\&= \sqrt{ 128 }\\&= 8\sqrt{ 2}\end{aligned}\)
因此，
\(\begin{aligned} AB + BC &= 2\sqrt{ 2} + 6\sqrt{ 2 }\\&= 8\sqrt{ 2 } \\&= AC \end{aligned}\)
既然如此，\(AB + BC = AC\)那么三个点是共线的。
- 确定以下几点是否共线。
- \(A(4,-1)\quad B(5,-2) \quad C(1,2)\)
- \(A(2,-2)\quad B(3,1)\quad C(2,1)\)