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CIE 9231 2024 June Paper 13 Q3

A Level / CIE / FP1

CIE 9231 2024 June Paper 13 Paper · Question 3

题目

Problem

The matrix M\mathbf{M} is given by M=(1201)(7001)\mathbf{M}=\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}7&0\\0&1\end{pmatrix}.

(a) The matrix M\mathbf{M} represents a sequence of two geometrical transformations in the xx-yy plane.

Give full details of each transformation, and make clear the order in which they are applied.

(4)

(b) Find the equations of the invariant lines, through the origin, of the transformation represented by M\mathbf{M}.

(5)

The triangle DEFDEF in the xx-yy plane is transformed by M\mathbf{M} onto triangle PQRPQR.

(c) Given that the area of triangle PQRPQR is 35 cm235\text{ cm}^2, find the area of triangle DEFDEF.

(2)
题目中文翻译

矩阵 M\mathbf{M} 定义为 M=(1201)(7001)\mathbf{M}=\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}7&0\\0&1\end{pmatrix}。

(a) 矩阵 M\mathbf{M} 表示在 xx-yy 平面上依次进行的两个几何变换。

详细说明每个变换,并明确指出它们的实施顺序。

(4)

(b) 求矩阵 M\mathbf{M} 所表示的变换的所有经过原点的不变直线方程。

(5)

xx-yy 平面上的三角形 DEFDEF 经矩阵 M\mathbf{M} 变换后得到三角形 PQRPQR。

(c) 已知三角形 PQRPQR 的面积为 35 cm235\text{ cm}^2,求三角形 DEFDEF 的面积。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

右侧矩阵先作用:先横向拉伸,再沿固定横轴作剪切,必须说明剪切中单位纵向点的像。

答题过程

展开

First apply a one-way stretch parallel to the xx-axis, scale factor 77.

Then apply a shear with the xx-axis fixed, taking (0,1)(0,1) to (2,1)(2,1).

(b)

解法一

思路

展开

设过原点直线为y=mx,要求像点的纵横坐标仍满足此关系,再检查竖直线。

答题过程

展开

The combined matrix is

M=(7201).\mathbf M=\begin{pmatrix}7&2\\0&1\end{pmatrix}.

A point (x,mx)(x,mx) maps to ((7+2m)x,mx)((7+2m)x,mx). For Y=mXY=mX,

m=m(7+2m)⟹2m(m+3)=0.m=m(7+2m)\quad\Longrightarrow\quad 2m(m+3)=0.

Thus m=0m=0 or m=−3m=-3. The corresponding horizontal factors are 77 and 11, both nonzero, so each line maps onto itself.

The vertical line is not invariant, since (0,y)(0,y) maps to (2y,y)(2y,y). Therefore

y=0andy=−3x.\boxed{y=0\quad\text{and}\quad y=-3x}.

(c)

解法一

思路

展开

面积乘以行列式绝对值7,因此原像面积是像面积除7。

答题过程

展开 Area⁡(PQR)= ∣det⁡M∣Area⁡(DEF)= 7Area⁡(DEF).\begin{align*} \operatorname{Area}(PQR)=&\,|\det\mathbf M|\operatorname{Area}(DEF)\\[4mm] =&\,7\operatorname{Area}(DEF). \end{align*}

Hence Area⁡(DEF)=35/7=5 cm2\boxed{\operatorname{Area}(DEF)=35/7=5\text{ cm}^2}.