| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding the equation of a line through two points for part (i), and finding the perpendicular bisector equation for part (ii). Both parts involve routine algebraic manipulation with no conceptual challenges or novel problem-solving required, making it easier than the average A-level question. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m\) of \(AB\) is \(-\frac{1}{2}\) oe; Eqn of \(AB\) is \(y = -\frac{1}{2}x + 7\); Let \(x = 3k\), \(y = k\) | B1, M1, M1 | Using \(A,B\) or \(C\) to get an equation; Using \(C\) or \(A,B\) in the equation |
| \(k = 2.8\) oe | A1 | |
| OR \(\frac{7-k}{0-3k} = \frac{3-k}{8-3k}\) | M1A1 | Using \(A,B\) & \(C\) to equate gradients |
| \(\to 20k = 56 \to k = 2.8\) | DM1A1 | Simplifies to a linear or 3 term quadratic \(= 0\) |
| OR \(\frac{7-k}{0-3k} = \frac{7-3}{0-8}\) | M1A1 | Using \(A,B\) and \(C\) to equate gradients |
| \(\to 20k = 56 \to k = 2.8\) | DM1A1 | Simplifies to a linear or 3 term quadratic \(= 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(4, 5)\) | B1 | anywhere in (ii) |
| Perpendicular gradient \(= 2\) | M1 | Use of \(m_1m_2=-1\) |
| Perp bisector has eqn \(y-5=2(x-4)\) | M1 | Forming eqn using their \(M\) and their "perpendicular \(m\)" |
| Let \(x=3k\), \(y=k\); \(k=\frac{3}{5}\) oe | A1 | |
| OR | ||
| \((0-3k)^2+(7-k)^2=(8-3k)^2+(3-k)^2\) | M1A1 | Use of Pythagoras |
| \(-14k+49=73-54k \rightarrow 40k=24 \rightarrow k=0.6\) | DM1A1 [4] | Simplifies to a linear or 3 term quadratic \(=0\) |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m$ of $AB$ is $-\frac{1}{2}$ oe; Eqn of $AB$ is $y = -\frac{1}{2}x + 7$; Let $x = 3k$, $y = k$ | B1, M1, M1 | Using $A,B$ or $C$ to get an equation; Using $C$ or $A,B$ in the equation |
| $k = 2.8$ oe | A1 | |
| **OR** $\frac{7-k}{0-3k} = \frac{3-k}{8-3k}$ | M1A1 | Using $A,B$ & $C$ to equate gradients |
| $\to 20k = 56 \to k = 2.8$ | DM1A1 | Simplifies to a linear or 3 term quadratic $= 0$ |
| **OR** $\frac{7-k}{0-3k} = \frac{7-3}{0-8}$ | M1A1 | Using $A,B$ and $C$ to equate gradients |
| $\to 20k = 56 \to k = 2.8$ | DM1A1 | Simplifies to a linear or 3 term quadratic $= 0$ |
# Question (ii) [Perpendicular Bisector]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(4, 5)$ | **B1** | anywhere in (ii) |
| Perpendicular gradient $= 2$ | **M1** | Use of $m_1m_2=-1$ |
| Perp bisector has eqn $y-5=2(x-4)$ | **M1** | Forming eqn using their $M$ and their "perpendicular $m$" |
| Let $x=3k$, $y=k$; $k=\frac{3}{5}$ oe | **A1** | |
| **OR** | | |
| $(0-3k)^2+(7-k)^2=(8-3k)^2+(3-k)^2$ | **M1A1** | Use of Pythagoras |
| $-14k+49=73-54k \rightarrow 40k=24 \rightarrow k=0.6$ | **DM1A1** [4] | Simplifies to a linear or 3 term quadratic $=0$ |
---8 Three points have coordinates $A ( 0,7 ) , B ( 8,3 )$ and $C ( 3 k , k )$. Find the value of the constant $k$ for which\\
(i) $C$ lies on the line that passes through $A$ and $B$,\\
(ii) $C$ lies on the perpendicular bisector of $A B$.
\hfill \mbox{\textit{CAIE P1 2016 Q8 [8]}}