AQA FP2 2010 June — Question 3 9 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2010
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeIntersection of two loci
DifficultyStandard +0.3 This is a standard Further Maths FP2 loci question requiring routine techniques: verifying a point lies on two loci (direct substitution), sketching a perpendicular bisector and a half-line from the origin, and shading a region. All steps are textbook exercises with no novel insight required, making it slightly easier than average even for Further Maths.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

3 Two loci, \(L _ { 1 }\) and \(L _ { 2 }\), in an Argand diagram are given by $$\begin{aligned} & L _ { 1 } : | z + 1 + 3 \mathrm { i } | = | z - 5 - 7 \mathrm { i } | \\ & L _ { 2 } : \arg z = \frac { \pi } { 4 } \end{aligned}$$
  1. Verify that the point represented by the complex number \(2 + 2 \mathrm { i }\) is a point of intersection of \(L _ { 1 }\) and \(L _ { 2 }\).
  2. Sketch \(L _ { 1 }\) and \(L _ { 2 }\) on one Argand diagram.
  3. Shade on your Argand diagram the region satisfying
    both $$| z + 1 + 3 i | \leqslant | z - 5 - 7 i |$$ and $$\frac { \pi } { 4 } \leqslant \arg z \leqslant \frac { \pi } { 2 }$$

Question 3:
Part (a):
AnswerMarks Guidance
For \(L_1\): \(2+2i+1+3i =
For \(L_2\): \(\arg(2+2i) = \arctan(1) = \frac{\pi}{4}\) ✓B1 Verifies \(L_2\)
Part (b):
AnswerMarks Guidance
\(L_1\) is perpendicular bisector of points \((-1,-3)\) and \((5,7)\); midpoint \((2,2)\), correctly drawnB1 Midpoint identified
Correct straight line for \(L_1\)B1 Correct perpendicular bisector
\(L_2\) is a half-line from origin at angle \(\frac{\pi}{4}\)B1 Correct half-line
Both lines intersecting at \((2,2)\)B1 Correct intersection
Fully correct diagramB1 All correct
Part (c):
AnswerMarks Guidance
Region on or to the right of \(L_1\) (i.e. \(z+1+3i \leq
Region between half-lines at \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\) (including boundaries)B1 Correct angular region
# Question 3:

## Part (a):
| For $L_1$: $|2+2i+1+3i| = |3+5i| = \sqrt{34}$; $|2+2i-5-7i| = |-3-5i| = \sqrt{34}$ ✓ | B1 | Verifies $L_1$ |
| For $L_2$: $\arg(2+2i) = \arctan(1) = \frac{\pi}{4}$ ✓ | B1 | Verifies $L_2$ |

## Part (b):
| $L_1$ is perpendicular bisector of points $(-1,-3)$ and $(5,7)$; midpoint $(2,2)$, correctly drawn | B1 | Midpoint identified |
|---|---|---|
| Correct straight line for $L_1$ | B1 | Correct perpendicular bisector |
| $L_2$ is a half-line from origin at angle $\frac{\pi}{4}$ | B1 | Correct half-line |
| Both lines intersecting at $(2,2)$ | B1 | Correct intersection |
| Fully correct diagram | B1 | All correct |

## Part (c):
| Region on or to the right of $L_1$ (i.e. $|z+1+3i| \leq |z-5-7i|$) | B1 | Correct side shaded |
|---|---|---|
| Region between half-lines at $\frac{\pi}{4}$ and $\frac{\pi}{2}$ (including boundaries) | B1 | Correct angular region |

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3 Two loci, $L _ { 1 }$ and $L _ { 2 }$, in an Argand diagram are given by

$$\begin{aligned}
& L _ { 1 } : | z + 1 + 3 \mathrm { i } | = | z - 5 - 7 \mathrm { i } | \\
& L _ { 2 } : \arg z = \frac { \pi } { 4 }
\end{aligned}$$

(a) Verify that the point represented by the complex number $2 + 2 \mathrm { i }$ is a point of intersection of $L _ { 1 }$ and $L _ { 2 }$.\\
(b) Sketch $L _ { 1 }$ and $L _ { 2 }$ on one Argand diagram.\\
(c) Shade on your Argand diagram the region satisfying\\
both

$$| z + 1 + 3 i | \leqslant | z - 5 - 7 i |$$

and

$$\frac { \pi } { 4 } \leqslant \arg z \leqslant \frac { \pi } { 2 }$$

\hfill \mbox{\textit{AQA FP2 2010 Q3 [9]}}