根据附件图片内容，讲解从 动点轨迹为直线的主从动 模型---**能力点 动点轨迹为直线的"主从联动"模型**
**方法总结**
| 特点 | 图示 | 结论 |
| :------------------------------------------------------------------- | :-------------------------------------------------------------------------------------------------------------------------------------------------------------------- | :----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| 定点 A, 动点 P 和 Q, $\angle PAQ = \alpha$, $\frac{AP}{AQ}$ 为定值 k, 点 P 在直线 BC 上运动 | [Diagram 1: Points A, Q, P. Line BC. P on BC. Q on AP. Arrow from P to C on BC.] [Diagram 2: Points A, Q, Q', P, P'. Line BC. P on BC. Q on AP. P' on BC. Q' on AP'. Arrow from P to P'. Line QQ' parallel to PP'.] [Diagram 3: Points A, Q, Q', P, P'. Line BC. P on BC. Q on line from A. P' on BC. Q' on line from A. $\angle PAQ = \alpha$, $\angle P'AQ' = \alpha$. Dashed line QQ'. Angle $\alpha$ between AQ and AQ'. Angle $\alpha$ between AP and AP'.] A, Q, P 三点共线 [under Diagram 2] A, Q, P 三点不共线 [under Diagram 3] | 当点 P 的轨迹在直线上时, 点 Q 的轨迹在直线上: ①P, Q 两点轨迹所在直线的夹角等于 $\alpha$; ②P, Q 两点轨迹长度之比等于 AP: AQ = k: 1 |
**方法应用**
**例** 如图, 在 $\triangle ABC$ 中, $BC=12$, M 为 BC 边上一点, 连接 AM, N 为线段 AM 上靠近点 M 的三等分点, 当点 M 从点 B 运动到点 C 时, 在图中作出动点 N 的运动轨迹, 并求出点 N 运动的路径长为____.
**关键点点拨**
三点共线时动点轨迹的确定: 结合平行线分线段成比例, 过从动点作主动点所在直线的平行线, 即为所求从动点运动轨迹所在直线.
**题图**
Diagram: A triangle ABC. Point M is on side BC. Point N is on line segment AM, closer to M. A line segment AM is drawn. A line segment connecting B and C is drawn. A line segment connecting A and B is drawn. A line segment connecting A and C is drawn. Points B, M, C are collinear. Points A, N, M are collinear.
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**Chart/Diagram Description:**
**Section: 方法总结 (Method Summary)**
* **Diagram 1:**
* Type: Geometric figure.
* Elements:
* Points: A, Q, P, B, C.
* Lines: Straight line BC. Line segment AP. Line segment AQ. Point P is on line BC. Point Q is on line segment AP.
* Arrows: An arrow points from P towards C along the line BC, indicating movement or direction.
* Labels: A, Q, P, B, C.
* Relationship: P is on the line BC. Q is on the line segment AP.
* **Diagram 2:**
* Type: Geometric figure illustrating translation/scaling.
* Elements:
* Points: A, Q, Q', P, P', B, C.
* Lines: Straight line BC. Line segment AP. Line segment AQ. Line segment AP'. Line segment AQ'. Dashed line segment QQ'. Solid line segment PP'.
* Arrows: An arrow points from P to P' along the line BC, indicating movement or direction. A dashed arrow points from Q to Q'.
* Labels: A, Q, Q', P, P', B, C. Text label "A, Q, P 三点共线" below the diagram.
* Relationship: P and P' are on line BC. Q is on AP, Q' is on AP'. QQ' appears parallel to PP'. A, Q, P are collinear.
* **Diagram 3:**
* Type: Geometric figure illustrating rotation and scaling.
* Elements:
* Points: A, Q, Q', P, P', B, C.
* Lines: Straight line BC. Line segments AP, AQ, AP', AQ'. Dashed line segments QQ' and PP'.
* Labels: A, Q, Q', P, P', B, C. Angle label $\alpha$ between AQ and AQ'. Angle label $\alpha$ between AP and AP'. Text label "A, Q, P 三点不共线" below the diagram.
* Relationship: P and P' are on line BC. Q is on a line from A, Q' is on a line from A. $\angle PAQ = \alpha$ and $\angle P'AQ' = \alpha$ implies a rotation around A. A, Q, P are not collinear.
**Section: 题图 (Problem Figure)**
* **Diagram:**
* Type: Geometric figure (Triangle).
* Elements:
* Points: A, B, C, M, N.
* Lines: Straight line segments forming triangle ABC. Point M is on the side BC. Line segment AM is drawn. Point N is on line segment AM.
* Labels: A, B, C, M, N. Text label "题图" below the diagram.
* Relationship: A, B, C form a triangle. M is on BC. N is on AM. N is described as a trisection point of AM closer to M in the problem text (not explicitly shown in the diagram itself, but consistent with N being between A and M). B, M, C are collinear. A, N, M are collinear.