CAIE P1 2015 November — Question 6 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeParameter from distance condition
DifficultyModerate -0.3 This is a straightforward coordinate geometry question requiring standard distance formula application and perpendicular bisector calculation. Part (i) involves setting up and solving a quadratic equation from AB=BC, while part (ii) requires finding midpoint, perpendicular gradient, and line intersection—all routine AS-level techniques with no novel problem-solving required.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors

6 Points \(A , B\) and \(C\) have coordinates \(A ( - 3,7 ) , B ( 5,1 )\) and \(C ( - 1 , k )\), where \(k\) is a constant.
  1. Given that \(A B = B C\), calculate the possible values of \(k\). The perpendicular bisector of \(A B\) intersects the \(x\)-axis at \(D\).
  2. Calculate the coordinates of \(D\).

Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(AB = 10\)B1
\(6^2 + (k-1)^2 = 10^2\)M1 Use of Pythagoras
\(k = -7\) and \(9\)A1 [3]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(m\) of \(AB = -\frac{3}{4}\), \(m\) perp \(= \frac{4}{3}\)B1 M1 B1 M1 Use of \(m_1 m_2 = -1\)
\(M = (1, 4)\)B1
Eqn \(y - 4 = \frac{4}{3}(x-1)\)M1 A1
Set \(y\) to \(0\), \(\rightarrow x = -2\)[5] Complete method leading to \(D\)
## Question 6:

### Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB = 10$ | **B1** | |
| $6^2 + (k-1)^2 = 10^2$ | **M1** | Use of Pythagoras |
| $k = -7$ and $9$ | **A1** [3] | |

### Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m$ of $AB = -\frac{3}{4}$, $m$ perp $= \frac{4}{3}$ | **B1 M1** | B1 M1 Use of $m_1 m_2 = -1$ |
| $M = (1, 4)$ | **B1** | |
| Eqn $y - 4 = \frac{4}{3}(x-1)$ | **M1 A1** | |
| Set $y$ to $0$, $\rightarrow x = -2$ | [5] | Complete method leading to $D$ |

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6 Points $A , B$ and $C$ have coordinates $A ( - 3,7 ) , B ( 5,1 )$ and $C ( - 1 , k )$, where $k$ is a constant.\\
(i) Given that $A B = B C$, calculate the possible values of $k$.

The perpendicular bisector of $A B$ intersects the $x$-axis at $D$.\\
(ii) Calculate the coordinates of $D$.

\hfill \mbox{\textit{CAIE P1 2015 Q6 [8]}}