| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Area with optimization or parameters |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry question requiring standard techniques: finding line equations, intercepts, and midpoints. Part (i) involves routine calculation of triangle area. Part (ii) requires finding a perpendicular line and verifying a midpoint property, but follows a clear algorithmic approach with no novel insight needed. Slightly above average due to the algebraic manipulation with parameter t and the two-part structure, but well within typical P1 expectations. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| (ii) | y−2t =−2(x−3t)(y+2x=8t) |
| Answer | Marks |
|---|---|
| This lies on the line y = x. | M1 |
| Answer | Marks |
|---|---|
| [4] | Unsimplified or equivalent forms |
Question 6:
--- 6 (i)
(ii) ---
6 (i)
(ii) | y−2t =−2(x−3t)(y+2x=8t)
Set x to 0 → B(0, 8t)
Set y to 0 → A(4t, 0)
→ Area = 16t²
1
m =
2
1
y−2t = (x−3t)(2y = x+t)
→
2
Set y to 0 → C (−t, 0)
Midpoint of CP is (t, t)
This lies on the line y = x. | M1
M1
A1
[3]
B1
M1
A1
A1
[4] | Unsimplified or equivalent forms
Attempt at both A and B, then using
cao
cao
Unsimplified or equivalent forms
co
correctly shown.The line with gradient $-2$ passing through the point $P(3t, 2t)$ intersects the $x$-axis at $A$ and the $y$-axis at $B$.
\begin{enumerate}[label=(\roman*)]
\item Find the area of triangle $AOB$ in terms of $t$. [3]
\end{enumerate}
The line through $P$ perpendicular to $AB$ intersects the $x$-axis at $C$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the mid-point of $PC$ lies on the line $y = x$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2015 Q6 [7]}}