| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.3 Part (i) is straightforward application of the gradient formula. Part (ii) requires using the distance formula and solving a quadratic equation, combining the linear constraint from (i) with the circle equation AP=AB. This is a standard multi-step coordinate geometry problem requiring algebraic manipulation but no novel insight—slightly easier than average due to clear structure and routine techniques. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| 8(i) | ( b−1 ) / ( a+1 )=2 | M1 |
| b=2a+3 CAO | A1 | Sub x=a , y=b → b=2a+3 |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(ii) | AB2 =112 +22 =125 oe | B1 |
| ( a+1 )2 +( b−1 )2 =125 | B1 FT | FT on their 125. |
| ( a+1 )2 +( 2a+2 )2 =125 | M1 | Sub from part (i) → quadratic eqn in a (or possibly in b → b2 −2b−99=0) |
| Answer | Marks | Guidance |
|---|---|---|
| ( 5 ) a2 +2a−24 =0→eg ( a−4 )( a+6 )=0 | M1 | Simplify and attempt to solve |
| a = 4 or −6 | A1 | |
| b = 11 or −9 | A1 | OR (4, 11), (‒6, ‒9) |
| Answer | Marks | Guidance |
|---|---|---|
| Total: | 6 | |
| Question | Answer | Marks |
Question 8:
--- 8(i) ---
8(i) | ( b−1 ) / ( a+1 )=2 | M1 | OR Equation of AP is y−1=2 ( x+1 ) → y=2x+3
b=2a+3 CAO | A1 | Sub x=a , y=b → b=2a+3
Total: | 2
--- 8(ii) ---
8(ii) | AB2 =112 +22 =125 oe | B1 | Accept AB = √125
( a+1 )2 +( b−1 )2 =125 | B1 FT | FT on their 125.
( a+1 )2 +( 2a+2 )2 =125 | M1 | Sub from part (i) → quadratic eqn in a (or possibly in b → b2 −2b−99=0)
( )
( 5 ) a2 +2a−24 =0→eg ( a−4 )( a+6 )=0 | M1 | Simplify and attempt to solve
a = 4 or −6 | A1
b = 11 or −9 | A1 | OR (4, 11), (‒6, ‒9)
If A0A0, SR1 for either (4, 11) or (‒6, ‒9)
Total: | 6
Question | Answer | Marks | Guidance$A(-1, 1)$ and $P(a, b)$ are two points, where $a$ and $b$ are constants. The gradient of $AP$ is 2.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $b$ in terms of $a$. [2]
\item $B(10, -1)$ is a third point such that $AP = AB$. Calculate the coordinates of the possible positions of $P$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q8 [8]}}