A-Level Mathematics May/June 2025 Q9(a): Three points P, Q and R have coordinates P(−13, 5), Q(5, 1) and R(2, k), where k is a c…

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  The coordinates of the points are P(−13, 5), Q(5, 1) and R(2, k). The angle PRQ is a right angle, which means the line segment PR is perpendicular to the line segment RQ.

  First, we find the gradients of PR and RQ. Gradient of PR, $m_{PR} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{k - 5}{2 - (-13)} = \frac{k - 5}{15}$

  Gradient of RQ, $m_{RQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{k - 1}{2 - 5} = \frac{k - 1}{-3}$

  Since PR is perpendicular to RQ, the product of their gradients is -1. $m_{PR} \times m_{RQ} = -1$

  $(\frac{k-5}{15}) \times (\frac{k-1}{-3}) = -1$

  $\frac{(k-5)(k-1)}{-45} = -1$

  $(k-5)(k-1) = 45$

  Expanding the brackets: $k^2 - k - 5k + 5 = 45$ $k^2 - 6k + 5 = 45$

  Rearranging to form a quadratic equation equal to zero: $k^2 - 6k - 40 = 0$

  To show that $k=10$ is a possible value, we can substitute it into the equation: $(10)^2 - 6(10) - 40 = 100 - 60 - 40 = 0$. This is true, so $k=10$ is a solution.

  To find the other possible value, we solve the quadratic equation, for example by factorising: $(k-10)(k+4) = 0$

  This gives two possible values for k: $k = 10$ or $k = -4$.

  Therefore, one possible value of k is 10, and the other possible value is -4.

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  How the marks are awarded

  • B1 — The first mark is for correctly finding the expression for at least one of the gradients. In the model answer, this is shown by writing $m_{PR} = \frac{k-5}{15}$ or $m_{RQ} = \frac{k-1}{-3}$.
  • M1 — The method mark is awarded for using the perpendicular gradient condition $m_1 \times m_2 = -1$, substituting the calculated gradients, and attempting to simplify to a quadratic form. This is seen where $(\frac{k-5}{15})(\frac{k-1}{-3}) = -1$ is simplified to $k^2 - 6k + 5 = 45$.
  • A1 — This accuracy mark is for obtaining the correct quadratic equation in the form $ax^2+bx+c=0$. The model answer shows the correct equation $k^2 - 6k - 40 = 0$.
  • A1 — The final accuracy mark is for correctly solving the quadratic to find both values of k. The answer correctly shows that $k=10$ is one value and finds the other value to be $k=-4$.

  Common mistakes

  • Incorrectly calculating a gradient, for example using $\frac{\Delta x}{\Delta y}$ or making a sign error with the negative coordinates, such as $2 - 13$ instead of $2 - (-13)$.
  • Using the wrong condition for perpendicular lines, such as $m_1 = m_2$ (for parallel lines) or $m_1 + m_2 = -1$, instead of the correct $m_1 \times m_2 = -1$.
  • Making an algebraic error when clearing the fractions, for example multiplying the right-hand side by $-15$ or $-3$ but not both, or making a sign error to get $(k-5)(k-1) = -45$.
  • Correctly finding the quadratic equation $k^2 - 6k - 40 = 0$ but then failing to factorise it correctly, leading to incorrect final values for $k$.

  Examiner tip: Remember that the product of the gradients of two perpendicular lines is -1; this is a fundamental coordinate geometry rule that frequently appears in questions involving right angles.

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  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

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