| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.3 Part (i) is a routine coordinate geometry exercise requiring midpoint calculation and perpendicular gradient (standard AS-level technique). Part (ii) involves substituting the line equation into the quadratic curve and using the distance formula with coordinates from solving a quadratic—straightforward multi-step work but no novel insight required. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Mid-point of \(AB = (3, 5)\) | B1 | Answers may be derived from simultaneous equations |
| Gradient of \(AB = 2\) | B1 | |
| Eqn of perp. bisector is \(y - 5 = -\frac{1}{2}(x-3) \rightarrow 2y = 13 - x\) | M1A1 | AG For M1 FT from mid-point and gradient of \(AB\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(-3x + 39 = 5x^2 - 18x + 19 \rightarrow (5)(x^2 - 3x - 4)(= 0)\) | M1 | Equate equations and form 3-term quadratic |
| \(x = 4\) or \(-1\) | A1 | |
| \(y = 4\frac{1}{2}\) or \(7\) | A1 | |
| \(CD^2 = 5^2 + 2\frac{1}{2}^2 \rightarrow CD = \sqrt{\dfrac{125}{4}}\) | M1A1 | Or equivalent integer fractions ISW |
## Question 6(i):
Mid-point of $AB = (3, 5)$ | B1 | Answers may be derived from simultaneous equations
Gradient of $AB = 2$ | B1 |
Eqn of perp. bisector is $y - 5 = -\frac{1}{2}(x-3) \rightarrow 2y = 13 - x$ | M1A1 | AG For M1 FT from mid-point and gradient of $AB$
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## Question 6(ii):
$-3x + 39 = 5x^2 - 18x + 19 \rightarrow (5)(x^2 - 3x - 4)(= 0)$ | M1 | Equate equations and form 3-term quadratic
$x = 4$ or $-1$ | A1 |
$y = 4\frac{1}{2}$ or $7$ | A1 |
$CD^2 = 5^2 + 2\frac{1}{2}^2 \rightarrow CD = \sqrt{\dfrac{125}{4}}$ | M1A1 | Or equivalent integer fractions ISW
---6 The points $A ( 1,1 )$ and $B ( 5,9 )$ lie on the curve $6 y = 5 x ^ { 2 } - 18 x + 19$.\\
(i) Show that the equation of the perpendicular bisector of $A B$ is $2 y = 13 - x$.\\
The perpendicular bisector of $A B$ meets the curve at $C$ and $D$.\\
(ii) Find, by calculation, the distance $C D$, giving your answer in the form $\sqrt { } \left( \frac { p } { q } \right)$, where $p$ and $q$ are integers.\\
\hfill \mbox{\textit{CAIE P1 2017 Q6 [9]}}