Plane transformations, coordinate reasoning and proof
ACT Topic 14 5:51 English narration · English + 中文 subtitles burned in
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Transcript
Rigid motions preserve lengths and angles.
刚体变换保持长度和角度。
A 90-degree anticlockwise rotation sends x, y to negative y, x about the origin.
绕原点逆时针旋转90度，将(x, y)映射为(-y, x)。
Distance is d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared.
距离公式为d等于根号下(x₂ - x₁)² + (y₂ - y₁)²。
Equal angles establish similarity, but congruence also fixes size.
相等的角可以确立相似性，但全等还固定了大小。
Each task gives its own data.
每个任务都有其自身的数据。
Do not copy the worked example’s numbers.
请勿照搬例题中的数值。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
使用前面页面的完整背景信息。
Rotate first, to 3 comma 2, then translate, to 7 comma 3.
先旋转，到（三，二），再平移，到（七，三）。
Translate first, to 6 comma negative 2, then rotate, to 2 comma 6.
先平移，到（六，负二），再旋转，到（二，六）。
The endpoints differ, so the order of transformations matters.
终点不同，所以变换的顺序很重要。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
The distance is the square root of 5 plus 1, squared, plus 10 minus 2, squared, which is 10.
距离等于五加一的平方加十减二的平方再开方，等于十。
The midpoint M is negative 1 plus 5 over 2, and 2 plus 10 over 2, which is 2 comma 6.
中点 M 的坐标是负一加五除以二，以及二加十除以二，即（二，六）。
The segment's slope is 8 over 6, or four thirds, so the perpendicular slope is negative three quarters.
线段的斜率是六分之八，即三分之四，所以垂线的斜率是负四分之三。
The bisector is y minus 6 equals negative three quarters times x minus 2.
垂直平分线是 y 减六等于负四分之三乘 x 减二。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
使用前面页面的完整背景信息。
The triangles are similar by equal angles, but not congruent, because the corresponding longest sides differ.
两个三角形因为对应角相等而相似，但并不全等，因为对应的最长边长度不同。
The length ratio is 15 over 9, or five thirds, and the larger-to-smaller area ratio is 25 over 9.
长度比是九分之十五，即三分之五，大三角形与小三角形的面积比是九分之二十五。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
Use the same compass radius, greater than half AB, to draw arcs centred at A and B meeting at P and Q on opposite sides.
用同一个圆规半径，且大于 AB 的一半，分别以 A 和 B 为圆心画弧，两弧在两侧交于 P 和 Q。
Draw PQ.
连接 PQ。
PA equals PB and QA equals QB by equal radii.
因为半径相等，PA 等于 PB，QA 等于 QB。
Thus both P and Q lie on the locus equidistant from A and B; PQ is the perpendicular bisector.
所以 P 和 Q 都在到 A、B 距离相等的轨迹上；PQ 就是垂直平分线。
Equivalently use congruent triangles and the common chord to establish perpendicularity and bisection.
也可以用全等三角形和公共弦来证明垂直和平分。
A ruler reading on one picture does not prove the general locus.
在一张图上用尺子量出的读数，并不能证明一般的轨迹。
SSS proves the two stated triangles congruent, fixing size as well as shape; Pythagoras also shows these particular triangles are right-angled.
边边边证明了所说的两个三角形全等，大小和形状都确定了；勾股定理还说明这两个特定的三角形是直角三角形。
Pause here and attempt the task before revealing the reasoning.
请先暂停，完成这道原创练习再继续。
Use the full context on the preceding pages.
请利用前几页的完整背景信息。
Parallelism gives the matching corresponding angle, 68 degrees.
由平行得到对应的同位角是六十八度。
The adjacent pair forms a straight angle, so the other angle is 180 minus 68, which is 112 degrees.
相邻的两个角组成平角，所以另一个角是一百八十减六十八，等于一百一十二度。
Equal corresponding angles require the stated parallel lines; the straight-angle sum uses adjacency on one line.
同位角相等需要题目给出的平行线；平角之和用的是同一直线上的相邻关系。
A similar-looking sketch without parallelism would not justify the first equality.
一张看起来相似但没有平行条件的草图，不能证明第一个等式。
Return to the task and explain why the method fits.
回到该任务，并解释为何此方法适用。
Original practice does not predict an official score.
原始练习无法预测正式评分。
Complete required live or recorded evidence separately.
需单独完成所需的现场或录制证据。
Pause and test yourself. Cover the worked answer, rebuild the answer to the last example in your own words, and name one condition that would change it before you check.
暂停一下，自我检测：遮住示范答案，用自己的话重做最后一个例子的答案，并在核对之前说出一个会改变结论的条件。