| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Perpendicular bisector of intersection points |
| Difficulty | Moderate -0.3 This is a standard simultaneous equations question combining linear and hyperbolic curves. Part (i) requires substitution, solving a quadratic, finding midpoint and perpendicular gradient—routine techniques. Part (ii) uses discriminant analysis, which is a common A-level method. The question is slightly easier than average as it follows predictable patterns with no novel insight required. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02q Use intersection points: of graphs to solve equations1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2y + x = 8 \to y(8-2y) = 6 \to 2y^2 - 8y + 6 = 0\) or \(x^2 - 8x + 12 = 0 \to\) (6, 1) and (2, 3) | M1, DM1, A1 | Complete elimination of x (or y); DM1 soln of quadratic; co |
| Midpoint \(M(4, 2)\); \(m = -\frac{1}{2}\); Perpendicular \(m = 2 \to y - 2 = 2(x - 4)\) | M1, M1, M1, A1 | for their 2 points; Uses \(m_1m_2 = -1\) to find perp. gradient; co unsimplified |
| (ii) \((k - 2y)y = 6 \to 2y^2 - ky + 6 = 0\) or \(x^2 - kx + 12 = 0\); Uses \(b^2 - 4ac\) (0); \(k^2 > 48 \to k < -\sqrt{48}\) and \(k > \sqrt{48}\) | M1, A1, A1, A1 | Any use of \(b^2 - 4ac\) on a quadratic \(= 0\); For \(\sqrt{48}\) on its own; All correct |
$2y + x = k$, $xy = 6$
**(i)** $2y + x = 8 \to y(8-2y) = 6 \to 2y^2 - 8y + 6 = 0$ or $x^2 - 8x + 12 = 0 \to$ (6, 1) and (2, 3) | M1, DM1, A1 | Complete elimination of x (or y); DM1 soln of quadratic; co
Midpoint $M(4, 2)$; $m = -\frac{1}{2}$; Perpendicular $m = 2 \to y - 2 = 2(x - 4)$ | M1, M1, M1, A1 | for their 2 points; Uses $m_1m_2 = -1$ to find perp. gradient; co unsimplified
**(ii)** $(k - 2y)y = 6 \to 2y^2 - ky + 6 = 0$ or $x^2 - kx + 12 = 0$; Uses $b^2 - 4ac$ (0); $k^2 > 48 \to k < -\sqrt{48}$ and $k > \sqrt{48}$ | M1, A1, A1, A1 | Any use of $b^2 - 4ac$ on a quadratic $= 0$; For $\sqrt{48}$ on its own; All correctThe equation of a line is $2y + x = k$, where $k$ is a constant, and the equation of a curve is $xy = 6$.
\begin{enumerate}[label=(\roman*)]
\item In the case where $k = 8$, the line intersects the curve at the points $A$ and $B$. Find the equation of the perpendicular bisector of the line $AB$. [6]
\item Find the set of values of $k$ for which the line $2y + x = k$ intersects the curve $xy = 6$ at two distinct points. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2012 Q10 [9]}}