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IAL 2024 June Q1

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 1

题目

Problem

The complex number z=x+iyz = x + iy satisfies the equation

∣z−3−4i∣=∣z+1+i∣\begin{align*} |z - 3 - 4\mathrm{i}| = |z + 1 + \mathrm{i}| \end{align*}

(a) Determine an equation for the locus of zz giving your answer in the form ax+by+c=0ax + by + c = 0 where aa , bb and cc are integers.

(3)

(b) Shade, on an Argand diagram, the region defined by

∣z−3−4i∣⩽∣z+1+i∣\begin{align*} |z - 3 - 4\mathrm{i}| \leqslant |z + 1 + \mathrm{i}| \end{align*}

You do not need to determine the coordinates of any intercepts on the coordinate axes.

(1)

解答

(a)

解法一

思路

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等式表示点 zz 到 3+4i3+4\mathrm{i} 和 −1−i-1-\mathrm{i} 的距离相等，所以轨迹是两点连线的垂直平分线。用模长平方展开最直接。

答题过程

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Let z=x+iyz=x+\mathrm{i}y. Then

∣z−3−4i∣= ∣(x−3)+i(y−4)∣∣z+1+i∣= ∣(x+1)+i(y+1)∣\begin{align*} |z-3-4\mathrm{i}| =&\, |(x-3)+\mathrm{i}(y-4)|\\[2mm] |z+1+\mathrm{i}| =&\, |(x+1)+\mathrm{i}(y+1)| \end{align*}

So

(x−3)2+(y−4)2= (x+1)2+(y+1)2x2−6x+9+y2−8y+16= x2+2x+1+y2+2y+1−6x−8y+25= 2x+2y+28x+10y−23= 0\begin{align*} (x-3)^2+(y-4)^2 =&\, (x+1)^2+(y+1)^2\\[4mm] x^2-6x+9+y^2-8y+16 =&\, x^2+2x+1+y^2+2y+1\\[4mm] -6x-8y+25 =&\, 2x+2y+2\\[4mm] 8x+10y-23 =&\,0 \end{align*}

Thus the locus is

8x+10y−23=0\begin{align*} \boxed{8x+10y-23=0} \end{align*}

解法二

思路

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也可以直接用垂直平分线。两点是 (3,4)(3,4) 和 (−1,−1)(-1,-1)。先求中点和原连线斜率，再取负倒数。

答题过程

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The two fixed points are

(3,4)and(−1,−1)\begin{align*} (3,4)\quad\text{and}\quad(-1,-1) \end{align*}

Their midpoint is

(3+(−1)2,4+(−1)2)=(1,32)\begin{align*} \left(\frac{3+(-1)}{2},\frac{4+(-1)}{2}\right) =\left(1,\frac32\right) \end{align*}

The gradient of the line joining the two points is

4−(−1)3−(−1)=54\begin{align*} \frac{4-(-1)}{3-(-1)}=\frac54 \end{align*}

So the perpendicular bisector has gradient −45-\frac45. Hence

y−32= −45(x−1)10y−15= −8x+88x+10y−23= 0\begin{align*} y-\frac32 =&\, -\frac45(x-1)\\[4mm] 10y-15 =&\, -8x+8\\[4mm] 8x+10y-23 =&\,0 \end{align*}

(b)

解法一

思路

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不等式表示点 zz 到 3+4i3+4\mathrm{i} 的距离不大于到 −1−i-1-\mathrm{i} 的距离，所以阴影在垂直平分线靠近 3+4i3+4\mathrm{i} 的一侧。可以用点 (3,4)(3,4) 判断是哪一边。

答题过程

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Draw the line

8x+10y−23=0\begin{align*} 8x+10y-23=0 \end{align*}

This line has negative gradient and positive yy-intercept.

Since the region is closer to 3+4i3+4\mathrm{i} than to −1−i-1-\mathrm{i}, shade the side of the line containing the point (3,4)(3,4). Equivalently, shade the region above the line.