| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry question requiring standard techniques: gradient formula, midpoint formula, and perpendicular gradient. Part (i) involves algebraic simplification to show independence from k (a nice touch but routine). Part (ii) combines these skills but follows a predictable method. Slightly easier than average due to being a standard textbook-style question with clear steps. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient \(m\) of \(AB = \frac{3k+5-(k+3)}{k+3-(-3k-1)}\) OE \(\left(= \frac{2k+2}{4k+4}\right) = \frac{1}{2}\) | M1A1 | Condone omission of brackets for M mark |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Mid-pt \(= \left[\frac{1}{2}(-3k-1+k+3),\ \frac{1}{2}(3k+5+k+3)\right] = \left(\frac{-2k+2}{2}, \frac{4k+8}{2}\right)\) SOI | B1B1 | B1 for \(\frac{-2k+2}{2}\), B1 for \(\frac{4k+8}{2}\) (ISW) or better, i.e. \((-k+1,\ 2k+4)\) |
| Gradient of perpendicular bisector is \(\frac{-1}{\textit{their } m}\) SOI, Expect \(-2\) | M1 | Could appear in subsequent equation and/or could be in terms of \(k\) |
| Equation: \(y-(2k+4) = -2\left[x-(-k+1)\right]\) OE | DM1 | Through *their* mid-point and with *their* \(\frac{-1}{m}\) (now numerical) |
| \(y + 2x = 6\) | A1 | Use of numerical \(k\) in (ii) throughout scores SC2/5 for correct answer |
| Total: 5 |
## Question 6(i):
| Gradient $m$ of $AB = \frac{3k+5-(k+3)}{k+3-(-3k-1)}$ OE $\left(= \frac{2k+2}{4k+4}\right) = \frac{1}{2}$ | M1A1 | Condone omission of brackets for M mark |
|---|---|---|
| **Total: 2** | | |
## Question 6(ii):
| Mid-pt $= \left[\frac{1}{2}(-3k-1+k+3),\ \frac{1}{2}(3k+5+k+3)\right] = \left(\frac{-2k+2}{2}, \frac{4k+8}{2}\right)$ SOI | B1B1 | B1 for $\frac{-2k+2}{2}$, B1 for $\frac{4k+8}{2}$ (ISW) or better, i.e. $(-k+1,\ 2k+4)$ |
|---|---|---|
| Gradient of perpendicular bisector is $\frac{-1}{\textit{their } m}$ SOI, Expect $-2$ | M1 | Could appear in subsequent equation and/or could be in terms of $k$ |
| Equation: $y-(2k+4) = -2\left[x-(-k+1)\right]$ OE | DM1 | Through *their* mid-point and with *their* $\frac{-1}{m}$ (now numerical) |
| $y + 2x = 6$ | A1 | Use of numerical $k$ in (ii) throughout scores SC2/5 for correct answer |
| **Total: 5** | | |
---6 The coordinates of points $A$ and $B$ are $( - 3 k - 1 , k + 3 )$ and $( k + 3,3 k + 5 )$ respectively, where $k$ is a constant ( $k \neq - 1$ ).\\
(i) Find and simplify the gradient of $A B$, showing that it is independent of $k$.\\
(ii) Find and simplify the equation of the perpendicular bisector of $A B$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q6 [7]}}