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九年级数学solution一般
题目
如图,在12×1212\times 12正方形网格中建立直角坐标系,每个小正方形的边长为11个单位长度,ABC\triangle ABC的三个顶点的坐标依次为:A(0,2)A\left(0,2\right),B(3,5)B\left(-3,5\right),C(2,2)C\left(-2,2\right).
(1)(1)ABC\triangle ABC以点AA为旋转中心旋转180180^{\circ},得到AB1C1\triangle AB_{1}C_{1},点BBCC的对应点分别为点B1B_{1}C1C_{1}请在网格图中画出AB1C1\triangle AB_{1}C_{1}.
(2)(2)ABC\triangle ABC平移至A2B2C2\triangle A_{2}B_{2}C_{2},其中点AABBCC的对应点分别为点A2A_{2}B2B_{2}C2C_{2},且点C2C_{2}的坐标为(2,4)\left(-2,-4\right),请在图中画出移后的A2B2C2\triangle A_{2}B_{2}C_{2}.
(3)(3)在第(1)、(2)小题基础上,若将AB1C1\triangle AB_{1}C_{1}绕某一点旋转可得到A2B2C2\triangle A_{2}B_{2}C_{2},则旋转中心的坐标为______.(直接写出答案)
(4)(4)xx轴上找出一点PP,使点PP到点B1B_{1},A2A_{2}的距离之和最小,直接写出点PP的坐标______.
知识点:坐标与图形变换——平移、作图——平移变换、三角形的面积章节:未标注

答案与解析

答案

(1)(1)如图,AB1C1\triangle AB_{1}C_{1}即可所求.
(2)(2)由题意得,ABC\triangle ABC向下平移66个单位长度得到A2B2C2\triangle A_{2}B_{2}C_{2}
如图,A2B2C2\triangle A_{2}B_{2}C_{2}即为所求.

(3)(3)连接AA2AA_{2}B1B2B_{1}B_{2}C1C2C_{1}C_{2},相交于点MM
AB1C1\triangle AB_{1}C_{1}绕点MM旋转180180^{\circ}得到A2B2C2\triangle A_{2}B_{2}C_{2}
\therefore旋转中心的坐标为(0,1)\left(0,-1\right).
故答案为:(0,1)\left(0,-1\right).
(4)(4)取点B1B_{1}关于xx轴的对称点B\’{B\’},连接A2B\’A_{2}{B\’}xx轴于点PP
此时点PP到点B1B_{1}A2A_{2}的距离之和为PB1+PA2=PB\’+PA2=A2B\’PB_{1}+PA_{2}=PB\’+PA_{2}=A_{2}{B\’},为最小值.
设直线A2B\’A_{2}{B\’}的解析式为y=kx+by=kx+b
A2(0,4),B\’(3,1)A_{2}(0,-4),{B\’}\left(3,1\right)代入,
{b=43k+b=1\left\{\begin{array}{l}b=-4\\ 3k+b=1\end{array}\right.
解得{k=53b=4\left\{\begin{array}{l}k=\frac{5}{3}\\ b=-4\end{array}\right.
则直线A2B\’A_{2}{B\’}的解析式为y=53x4y=\frac{5}{3}x-4
y=0y=0,得x=125x=\frac{12}{5}
\thereforePP的坐标为(1250)(\frac{12}{5},0).
故答案为:(1250)(\frac{12}{5},0).

解析

(1)(1)如图,AB1C1\triangle AB_{1}C_{1}即可所求.
(2)(2)由题意得,ABC\triangle ABC向下平移66个单位长度得到A2B2C2\triangle A_{2}B_{2}C_{2}
如图,A2B2C2\triangle A_{2}B_{2}C_{2}即为所求.

(3)(3)连接AA2AA_{2}B1B2B_{1}B_{2}C1C2C_{1}C_{2},相交于点MM
AB1C1\triangle AB_{1}C_{1}绕点MM旋转180180^{\circ}得到A2B2C2\triangle A_{2}B_{2}C_{2}
\therefore旋转中心的坐标为(0,1)\left(0,-1\right).
故答案为:(0,1)\left(0,-1\right).
(4)(4)取点B1B_{1}关于xx轴的对称点B\’{B\’},连接A2B\’A_{2}{B\’}xx轴于点PP
此时点PP到点B1B_{1}A2A_{2}的距离之和为PB1+PA2=PB\’+PA2=A2B\’PB_{1}+PA_{2}=PB\’+PA_{2}=A_{2}{B\’},为最小值.
设直线A2B\’A_{2}{B\’}的解析式为y=kx+by=kx+b
A2(0,4),B\’(3,1)A_{2}(0,-4),{B\’}\left(3,1\right)代入,
{b=43k+b=1\left\{\begin{array}{l}b=-4\\ 3k+b=1\end{array}\right.
解得{k=53b=4\left\{\begin{array}{l}k=\frac{5}{3}\\ b=-4\end{array}\right.
则直线A2B\’A_{2}{B\’}的解析式为y=53x4y=\frac{5}{3}x-4
y=0y=0,得x=125x=\frac{12}{5}
\thereforePP的坐标为(1250)(\frac{12}{5},0).
故答案为:(1250)(\frac{12}{5},0).

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