Question 7 - A-Level Maths

Edexcel F2 2014 June — Question 7 7 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2014
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex Numbers Argand & Loci
Type	Intersection of two loci
Difficulty	Standard +0.8 This question requires finding two loci (both perpendicular bisectors), determining their Cartesian equations, then solving simultaneously. While each locus is standard Further Maths content, the intersection requires algebraic manipulation and exact coordinate calculation, making it moderately challenging but within typical F2 scope.
Spec	4.02k Argand diagrams: geometric interpretation 4.02o Loci in Argand diagram: circles, half-lines

7. The point $P$ represents a complex number $z$ on an Argand diagram, where $$| z + 1 | = | 2 z - 1 |$$ and the point $Q$ represents a complex number $w$ on the Argand diagram, where $$| w | = | w - 1 + \mathrm { i } |$$ Find the exact coordinates of the points where the locus of $P$ intersects the locus of $Q$.

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Question 7:

Answer	Marks	Guidance
\(P:	z+1	=
Answer/Working	Mark	Guidance
\(z = x+iy \Rightarrow \	x+iy+1\	= \
$(x+1)^2 + y^2 = (2x-1)^2 + (2y)^2$	M1A1	M1: Correct use of Pythagoras. A1: Any correct equation
\(w = x+iy \Rightarrow \	x+iy\	= \
$x^2 + y^2 = (x-1)^2 + (y+1)^2 \ (\Rightarrow y = x-1)$	M1A1	M1: Correct use of Pythagoras. Allow with $u$ and $v$ instead of $x$ and $y$. A1: Any correct equation. Must have $x$ and $y$ now
Alternative: identifies perpendicular bisector of $(0,0)$ and $(1,-1)$; $y = x-1$	M1A1
$(x-1)^2 + (x-1)^2 = 1$ or $y^2 + y^2 = 1$	M1	Attempt to solve simultaneously to obtain an equation in one variable and get to $x=\ldots$ or $y=\ldots$
$x = 1+\frac{1}{\sqrt{2}}, 1-\frac{1}{\sqrt{2}}$ or $y = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$	A1	Both (oe)
$\left(1+\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(1-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$	A1	Both (oe). Pairs must be clearly identifiable but coordinate brackets not needed

Total: 7

## Question 7:

$P: |z+1| = |2z-1|$, $Q: |w| = |w-1+i|$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = x+iy \Rightarrow \|x+iy+1\| = \|2(x+iy)-1\|$ | | |
| $(x+1)^2 + y^2 = (2x-1)^2 + (2y)^2$ | M1A1 | M1: Correct use of Pythagoras. A1: Any correct equation |
| $w = x+iy \Rightarrow \|x+iy\| = \|x+iy-1+i\|$ | | |
| $x^2 + y^2 = (x-1)^2 + (y+1)^2 \ (\Rightarrow y = x-1)$ | M1A1 | M1: Correct use of Pythagoras. Allow with $u$ and $v$ instead of $x$ and $y$. A1: Any correct equation. Must have $x$ and $y$ now |
| Alternative: identifies perpendicular bisector of $(0,0)$ and $(1,-1)$; $y = x-1$ | M1A1 | |
| $(x-1)^2 + (x-1)^2 = 1$ or $y^2 + y^2 = 1$ | M1 | Attempt to solve simultaneously to obtain an equation in one variable and get to $x=\ldots$ or $y=\ldots$ |
| $x = 1+\frac{1}{\sqrt{2}}, 1-\frac{1}{\sqrt{2}}$ or $y = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$ | A1 | Both (oe) |
| $\left(1+\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(1-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$ | A1 | Both (oe). Pairs must be clearly identifiable but coordinate brackets not needed |

**Total: 7**

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7. The point $P$ represents a complex number $z$ on an Argand diagram, where

$$| z + 1 | = | 2 z - 1 |$$

and the point $Q$ represents a complex number $w$ on the Argand diagram, where

$$| w | = | w - 1 + \mathrm { i } |$$

Find the exact coordinates of the points where the locus of $P$ intersects the locus of $Q$.\\

\hfill \mbox{\textit{Edexcel F2 2014 Q7 [7]}}

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Q1 6 Q2 7 Q3 5 Q4 7 Q5 12 Q6 9 Q7 7 Q8 12 Q9 10