數與座標系/9 平面座標系-範例

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  • 已知：\left\{ \begin{align} & L_{1}:x+ay+b=0 \\ & L_{2}:x+cy+d=0 \\ \end{align} \right.，下列何者為真？
    • (A)a>0\, (B)b>0\, (C)c>0\, (D)d>0\, (E)a>c\,


  • 範例22
    • 已知：ΔABC之三頂點座標為：\left\{ \begin{align} & A=\left( 3,3 \right) \\ & B=\left( -1,-5 \right) \\ & C=(6,0) \\ \end{align} \right.
      1. 求解：ΔABC之重心座標
      2. 求解：ΔABC之外心座標
      3. 求解：ΔABC之垂心座標
  • 範例23


範例21


  • 已知：\left\{ \begin{align} & L_{1}:x+ay+b=0 \\ & L_{2}:x+cy+d=0 \\ \end{align} \right.，下列何者為真？
    • (A)a>0\, (B)b>0\, (C)c>0\, (D)d>0\, (E)a>c\,

答案

  • (D)(E)

詳解

  1. \left\{ \begin{align} & L_{1}:x+ay+b=0 \\ & L_{2}:x+cy+d=0 \\ \end{align} \right.
  2. L1的斜率大於0\,，故：m_{1}=-\frac{1}{a}>0\Rightarrow -\frac{1}{a}>0\Rightarrow a<0\#
  3. L1的x\,軸截距大於0\,，故：-b>0\Rightarrow b<0\#
  4. L1的斜率大於0\,，故，m_{2}=-\frac{1}{c}>0\Rightarrow -\frac{1}{c}>0\Rightarrow c<0\#
  5. L2的x\,軸截距小於0\,，故：-d<0\Rightarrow d>0\#
  6. \left\{ \begin{align} & \left\{ \begin{align} & a<0\# \\ & c<0\# \\ \end{align} \right.\Rightarrow ac>0 \\ & m_{1}>m_{2} \\ \end{align} \right.\Rightarrow -\frac{1}{a}>-\frac{1}{c}\Rightarrow -\frac{1}{a}\times ac>-\frac{1}{c}\times ac\Rightarrow -c>-a\Rightarrow a>c\#

類題1

已知

求解

答案

詳解

範例22

  • 已知：ΔABC之三頂點座標為：\left\{ \begin{align} & A=\left( 3,3 \right) \\ & B=\left( -1,-5 \right) \\ & C=(6,0) \\ \end{align} \right.
  • 求解：
    1. 重心座標
    2. 外心座標
    3. 垂心座標

答案

  1. \left( \frac{8}{3},-\frac{2}{3} \right)
  2. \left( \frac{5}{3},-\frac{4}{3} \right)
  3. \left( \frac{14}{3},\frac{2}{3} \right)

詳解

1重心座標

G=\frac{A+B+C}{3}=\left( \frac{3-1+6}{3},\frac{3-5+0}{3} \right)=\left( \frac{8}{3},-\frac{2}{3} \right)\#

2外心座標

  1. \left\{ \begin{align} & D=\frac{A+B}{2}=\left( \frac{3-1}{2},\frac{3-5}{2} \right)=\left( 1,-1 \right) \\ & E=\frac{B+C}{2}=\left( \frac{-1+6}{2},\frac{-5+0}{2} \right)=\left( \frac{5}{2},-\frac{5}{2} \right) \\ & F=\frac{A+C}{2}=\left( \frac{3+6}{2},\frac{3+0}{2} \right)=\left( \frac{9}{2},\frac{3}{2} \right) \\ \end{align} \right.
  2. \left\{ \begin{align} & \text{slope}\left( \overrightarrow{DG} \right)=-\frac{1}{\text{slope}\left( \overrightarrow{AB} \right)}=-\frac{3+1}{3+5}=-\frac{1}{2} \\ & \text{slope}\left( \overrightarrow{EG} \right)=-\frac{1}{\text{slope}\left( \overrightarrow{BC} \right)}=-\frac{6+1}{0+5}=-\frac{7}{5} \\ & \text{slope}\left( \overrightarrow{FG} \right)=-\frac{1}{\text{slope}\left( \overrightarrow{AC} \right)}=-\frac{6-3}{0-3}=1 \\ \end{align} \right.
  3. \left\{ \begin{align} & \overrightarrow{DG}:\frac{y+1}{x-1}=-\frac{1}{2}\Rightarrow 2y+2=-x+1 \\ & \overrightarrow{EG}:\frac{y+\frac{5}{2}}{x-\frac{5}{2}}=-\frac{7}{5}\Rightarrow 5y+\frac{25}{2}=-7x+\frac{35}{2} \\ & \overrightarrow{FG}:\frac{y-\frac{3}{2}}{x-\frac{9}{2}}=1\Rightarrow y-\frac{3}{2}=x-\frac{9}{2}\Rightarrow 2y-3=2x-9 \\ \end{align} \right.
  4. \left\{ \begin{align} & \overrightarrow{DG}:2y+2=-x+1 \\ & \overrightarrow{FG}:2y-3=2x-9 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & x=\frac{5}{3} \\ & y=-\frac{4}{3} \\ \end{align} \right.\Rightarrow G=\left( \frac{5}{3},-\frac{4}{3} \right)\#

3垂心座標

  1. \left\{ \begin{align} & \text{slope}\left( \overrightarrow{DC} \right)=-\frac{1}{\text{slope}\left( \overrightarrow{AB} \right)}=-\frac{3+1}{3+5}=-\frac{1}{2} \\ & \text{slope}\left( \overrightarrow{EA} \right)=-\frac{1}{\text{slope}\left( \overrightarrow{BC} \right)}=-\frac{6+1}{0+5}=-\frac{7}{5} \\ & \text{slope}\left( \overrightarrow{FB} \right)=-\frac{1}{\text{slope}\left( \overrightarrow{AC} \right)}=-\frac{6-3}{0-3}=1 \\ \end{align} \right.
  2. \left\{ \begin{align} & \overrightarrow{DC}:\frac{y-0}{x-6}=-\frac{1}{2}\Rightarrow 2y=-x+6 \\ & \overrightarrow{EA}:\frac{y-3}{x-3}=-\frac{7}{5}\Rightarrow 5y-15=-7x+21 \\ & \overrightarrow{FB}:\frac{y+5}{x+1}=1\Rightarrow y+5=x+1 \\ \end{align} \right.
  3. \left\{ \begin{align} & \overrightarrow{DC}:2y=-x+6 \\ & \overrightarrow{FB}:y+5=x+1 \\ \end{align} \right.\Rightarrow 3y+5=7\Rightarrow \left\{ \begin{align} & y=\frac{2}{3} \\ & x=\frac{14}{3} \\ \end{align} \right.\Rightarrow G=\left( \frac{14}{3},\frac{2}{3} \right)\#

類題1

已知

求解

答案

詳解

範例23

  • 已知：直線 L\, 通過點 (-4,1)\, ，且與兩座標軸圍成之面積為 1\, ，求：直線 L\, 之方程式

答案

  • L:\left\{ \begin{align} & x+8y=4 \\ & x+2y=-2 \\ \end{align} \right.

詳解


  1. 設直線 L\, 之方程式為 \frac{x}{a}+\frac{y}{b}=1
  2. 因通過點 (-4,1)\, ，故 \frac{-4}{a}+\frac{1}{b}=1
  3. 又因與座標軸圍成面積為 1\, ，故 \frac{\left| ab \right|}{2}=1\Rightarrow \left| ab \right|=\pm 2
    • 若 ab=2\, ，則
      \left\{ \begin{align} & ab=2\Rightarrow b=\frac{2}{a} \\ & \frac{-4}{a}+\frac{1}{b}=1\Rightarrow \frac{-4}{a}+\frac{a}{2}=1\Rightarrow -8+a^{2}=2a \\ \end{align} \right.
      \Rightarrow a^{2}-2a-8=0\Rightarrow a=\frac{2\pm \sqrt{4+32}}{2}=\frac{2\pm 6}{2}=4,or,-2
      \Rightarrow b=\frac{2}{4},or,\frac{2}{-2}\Rightarrow b=\frac{1}{2},or,-1
      \Rightarrow \left\{ \begin{align} & \frac{x}{4}+\frac{y}{\frac{1}{2}}=1\Rightarrow x+8y=4\# \\ & \frac{x}{-2}+\frac{y}{-1}=1\Rightarrow x+2y=-2\# \\ \end{align} \right.
    • 若 ab=-2\, ，則
      \left\{ \begin{align} & ab=-2\Rightarrow b=-\frac{2}{a} \\ & \frac{-4}{a}+\frac{1}{b}=1\Rightarrow \frac{-4}{a}-\frac{a}{2}=1\Rightarrow -8-a^{2}=2a \\ \end{align} \right.
      \Rightarrow a^{2}+2a+8=0\Rightarrow a=\frac{-2\pm \sqrt{4-32}}{2}\notin R\left( N/A \right)

範例23類題1

已知

求解

答案

詳解

範例23類題2

已知

求解

答案

詳解

範例23類題3

已知

求解

答案

詳解

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