WJEC Further Unit 1 2019 June — Question 6 3 marks

Exam BoardWJEC
ModuleFurther Unit 1 (Further Unit 1)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypePerpendicular bisector locus
DifficultyModerate -0.5 This is a standard locus question requiring students to translate a modulus equality into Cartesian form. The method is routine: substitute z = x + iy, expand both moduli using √[(x-a)² + (y-b)²], square both sides, and simplify. The algebra is straightforward and leads directly to a linear equation. This is easier than average as it's a textbook exercise with a well-practiced technique and minimal problem-solving required.
Spec4.02o Loci in Argand diagram: circles, half-lines

6. The complex number \(z\) is represented by the point \(P ( x , y )\) in an Argand diagram. Given that $$| z - 1 | = | z - 2 \mathrm { i } |$$ show that the locus of \(P\) is a straight line.

AnswerMarks Guidance
\(z - 1 =
$|z - 1| = |z - 2i|$, $|x + iy - 1| = |x + iy - 2i|$, $|(x-1) + iy| = |x + i(y-2)|$, $\sqrt{(x-1)^2 + y^2} = \sqrt{x^2 + (y-2)^2}$, $x^2 - 2x + 1 + y^2 = x^2 + y^2 - 4y + 4$, $-2x + 1 = -4y + 4$, $4y = 2x + 3$ which is a straight line oe | M1, m1, A1 |

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6. The complex number $z$ is represented by the point $P ( x , y )$ in an Argand diagram.

Given that

$$| z - 1 | = | z - 2 \mathrm { i } |$$

show that the locus of $P$ is a straight line.\\

\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q6 [3]}}