| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question testing basic skills: finding a perpendicular line (negative reciprocal of gradient) and calculating distance using the midpoint and distance formula. Both parts are routine textbook exercises requiring only direct application of standard formulas with no problem-solving or insight needed. |
| Spec | 1.03a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) gradient of perpendicular \(= -\frac{1}{2}\) soi | B1 | |
| \(y - 1 = -\frac{1}{2}(x - 3)\) | B1 | |
| [2] | ||
| (ii) \(C = (-9, 6)\) | B1 | soi in (i) or (ii) |
| \(AC^2 = [3 - (-9)]^2 + [1 - 6]^2\) (ft on their C) | M1 | OR \(AB^2 = [3 - (-21)]^2 + [1 - 11]^2\) M1 |
| \(AC = 13\) | A1 | \(AB = 26\) A1 |
| [3] | \(AC = 13\) A1 |
**(i)** gradient of perpendicular $= -\frac{1}{2}$ soi | B1 |
$y - 1 = -\frac{1}{2}(x - 3)$ | B1 |
| | [2]
**(ii)** $C = (-9, 6)$ | B1 | soi in (i) or (ii)
$AC^2 = [3 - (-9)]^2 + [1 - 6]^2$ (ft on their C) | M1 | OR $AB^2 = [3 - (-21)]^2 + [1 - 11]^2$ M1
$AC = 13$ | A1 | $AB = 26$ A1
| | [3] | $AC = 13$ A13 The point $A$ has coordinates $( 3,1 )$ and the point $B$ has coordinates $( - 21,11 )$. The point $C$ is the mid-point of $A B$.\\
(i) Find the equation of the line through $A$ that is perpendicular to $y = 2 x - 7$.\\
(ii) Find the distance $A C$.
\hfill \mbox{\textit{CAIE P1 2013 Q3 [5]}}