OCR MEI AS Paper 1 Specimen — Question 6 4 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeVector between two points
DifficultyModerate -0.3 This is a straightforward coordinate geometry problem using vectors. Students need to express C as (x, 1), apply the equal distance condition |AC| = |BC|, and solve a simple equation. It requires basic vector subtraction and magnitude calculation but involves minimal problem-solving beyond setting up the standard distance formula. Slightly easier than average due to the direct approach and limited steps.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10f Distance between points: using position vectors
6 Two points, \(A\) and \(B\), have position vectors \(\mathbf { a } = \mathbf { i } - 3 \mathbf { j }\) and \(\mathbf { b } = 4 \mathbf { i } + 3 \mathbf { j }\).
The point C lies on the line \(y = 1\). The lengths of the line segments AC and BC are equal. Determine the position vector of \(C\).
Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
Midpoint of \(AB = \left(\frac{5}{2}, 0\right)\)M1
Gradient of perpendicular to \(AB = -\frac{1}{2}\)M1
Perpendicular bisector equation is \(y = -\frac{1}{2}\left(x - \frac{5}{2}\right)\)A1
Position vector is \(\frac{1}{2}\mathbf{i} + \mathbf{j}\)A1
Alternative method:
AnswerMarks Guidance
AnswerMarks Guidance
Suppose C has position vector \(\mathbf{c} = p\mathbf{i} + \mathbf{j}\)M1
\(\overrightarrow{AC} = (p-1)\mathbf{i} + 4\mathbf{j}\) or \(\AC\ ^2 = (p-1)^2 + 4^2\)
\(\overrightarrow{BC} = (p-4)\mathbf{i} - 2\mathbf{j}\) or \(\BC\ ^2 = (p-4)^2 + 2^2\)
\((p-1)^2 + 4^2 = (p-4)^2 + 2^2\)A1 soi
Position vector is \(\frac{1}{2}\mathbf{i} + \mathbf{j}\)A1 
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoint of $AB = \left(\frac{5}{2}, 0\right)$ | M1 | |
| Gradient of perpendicular to $AB = -\frac{1}{2}$ | M1 | |
| Perpendicular bisector equation is $y = -\frac{1}{2}\left(x - \frac{5}{2}\right)$ | A1 | |
| Position vector is $\frac{1}{2}\mathbf{i} + \mathbf{j}$ | A1 | |

**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Suppose C has position vector $\mathbf{c} = p\mathbf{i} + \mathbf{j}$ | M1 | |
| $\overrightarrow{AC} = (p-1)\mathbf{i} + 4\mathbf{j}$ or $\|AC\|^2 = (p-1)^2 + 4^2$ | M1 | |
| $\overrightarrow{BC} = (p-4)\mathbf{i} - 2\mathbf{j}$ or $\|BC\|^2 = (p-4)^2 + 2^2$ | M1 | |
| $(p-1)^2 + 4^2 = (p-4)^2 + 2^2$ | A1 | soi |
| Position vector is $\frac{1}{2}\mathbf{i} + \mathbf{j}$ | A1 | |

---
6 Two points, $A$ and $B$, have position vectors $\mathbf { a } = \mathbf { i } - 3 \mathbf { j }$ and $\mathbf { b } = 4 \mathbf { i } + 3 \mathbf { j }$.\\
The point C lies on the line $y = 1$. The lengths of the line segments AC and BC are equal. Determine the position vector of $C$.

\hfill \mbox{\textit{OCR MEI AS Paper 1  Q6 [4]}}
This paper (12 questions)
View full paper
Q1 2 Q2 3 Q3 3 Q4 3 Q5 5 Q6 4 Q7 4 Q8 11 Q9 8 Q10 12 Q11 6 Q12 9