Edexcel IAL P1 June 2025 question paper and mark scheme (WMA11/01)

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1. Algebra · 7 marks

   Solve the simultaneous equations $x - y = 2$ and $x^2 - xy + y^2 = 7$.

   Worked answer

   M1 Rearranges the linear equation to make $y$ the subject: $y = x - 2$.

   1. M1 Substitutes this expression into the quadratic equation: $x^2 - x(x - 2) + (x - 2)^2 = 7$.
   2. M1 Expands the brackets correctly: $x^2 - x^2 + 2x + (x^2 - 4x + 4) = 7$.
   3. A1 Simplifies to a 3-term quadratic equation: $x^2 - 2x - 3 = 0$.
   4. M1 Factorises the quadratic: $(x - 3)(x + 1) = 0$.
   5. A1 Finds the $x$-values: $x = 3$ and $x = -1$.
   6. A1 Substitutes back into the linear equation to find the corresponding $y$-values, giving the final solution pairs $x = 3, y = 1$ and $x = -1, y = -3$.

2. Coordinate Geometry · 8 marks

   Three points define the vertices of a triangle: \(A(-2, -1)\), \(B(4, 7)\), and \(C(8, 4)\).

   1. By calculating the gradients of \([AB]\) and \([BC]\), prove that triangle \(ABC\) is a right-angled triangle.

   2. Calculate the exact lengths of \([AB]\) and \([BC]\).

   3. Hence, find the exact area of the triangle \(ABC\).

   Worked answer

   1. Gradient \(m_{AB} = \frac{7 - (-1)}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}\). A1
      Gradient \(m_{BC} = \frac{4 - 7}{8 - 4} = \frac{-3}{4}\). A1
      Since \(m_{AB} \times m_{BC} = \frac{4}{3} \times -\frac{3}{4} = -1\), \([AB]\) and \([BC]\) are perpendicular. M1
      Thus, triangle \(ABC\) has a right angle at \(B\). B1

   2. Length \(AB = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{36 + 64} = \sqrt{100} = 10\). A1
      Length \(BC = \sqrt{(8 - 4)^2 + (4 - 7)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\). A1

   3. Since the triangle is right-angled at \(B\), \(AB\) and \(BC\) act as the base and height. M1
      Area \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 5 = 25\) sq units. A1

3. Trigonometry · 5 marks

   Triangle $ABC$ has sides $AB = 4$ cm and $BC = 5$ cm, with angle $ABC = 60^\circ$. Triangle $ACD$ shares side $AC$ and has a right angle at $C$. If $CD = 7$ cm, find the exact length of $AD$.

   Worked answer

   1. M1 for using the cosine rule in triangle ABC to find $AC^2$: $AC^2 = 4^2 + 5^2 - 2(4)(5)\cos 60^\circ$.
   2. M1 for evaluating to get $AC^2 = 16 + 25 - 40(0.5) = 21$.
   3. M1 for recognising that triangle $ACD$ is right-angled at $C$, so Pythagoras' theorem applies: $AD^2 = AC^2 + CD^2$.
   4. M1 for substituting the known values: $AD^2 = 21 + 7^2 = 21 + 49 = 70$.
   5. A1 for finding $AD = \sqrt{70}$ cm.

More on these topics: Algebra · Coordinate Geometry · Trigonometry

P1 topics this paper can test

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• Algebra
• Coordinate Geometry
• Trigonometry
• Differentiation
• Integration
• Graphs and Transformations

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What the examiners said about this paper

Summarised in our own words from the board's examiner report (Examiner Insights).

• WMA11/01: Calculator warning ignored (Q2); Indices (Q10); Radians (Q6). Official examiner report ↗

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Formula booklet

Mathematical Formulae and Statistical Tables (International AS/A Level Mathematics) · Pearson Edexcel · For the 2018 IAL specification (current)