| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Midpoint of line segment |
| Difficulty | Moderate -0.3 This is a straightforward multi-part coordinate geometry question requiring standard techniques: solving simultaneous equations to find intersection points, using the midpoint formula, differentiation to find parallel tangent, and distance formula. All steps are routine AS-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 4x + 7 = 9 - 3x \rightarrow x^2 - x - 2 = 0\) | M1 | Complete elimination of \(y\) (or \(x\)) |
| Solution of this \(x = 2\) or \(-1\) | DM1 | Correct solution of eqn \(= 0\) |
| \(\rightarrow (2, 3)\) and \((-1, 12)\) | A1 | All 4 values needed |
| Mid point is \(M\,(\frac{1}{2},\, 7\frac{1}{2})\) | A1 | [4] Beware fortuitous ans. Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(dy/dx = 2x - 4\) | B1 | Co |
| Equate to \(m\) of line \((-3)\) + solution | M1 | Equates \(dy/dx\) to constant \(m\), \(m \neq 0\). Must have calculus – not for perp \(m\) |
| \(\rightarrow (\frac{1}{2},\, 5\frac{1}{4})\) | A1 | [3] Co |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distance \(= 2\frac{1}{4}\) | \(\text{B1}\sqrt{}\) | [1] For distance between "his" points |
## Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 4x + 7 = 9 - 3x \rightarrow x^2 - x - 2 = 0$ | M1 | Complete elimination of $y$ (or $x$) |
| Solution of this $x = 2$ or $-1$ | DM1 | Correct solution of eqn $= 0$ |
| $\rightarrow (2, 3)$ and $(-1, 12)$ | A1 | All 4 values needed |
| Mid point is $M\,(\frac{1}{2},\, 7\frac{1}{2})$ | A1 | [4] Beware fortuitous ans. Answer given |
## Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $dy/dx = 2x - 4$ | B1 | Co |
| Equate to $m$ of line $(-3)$ + solution | M1 | Equates $dy/dx$ to constant $m$, $m \neq 0$. Must have calculus – not for perp $m$ |
| $\rightarrow (\frac{1}{2},\, 5\frac{1}{4})$ | A1 | [3] Co |
## Question 5(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance $= 2\frac{1}{4}$ | $\text{B1}\sqrt{}$ | [1] For distance between "his" points |5 The equation of a curve is $y = x ^ { 2 } - 4 x + 7$ and the equation of a line is $y + 3 x = 9$. The curve and the line intersect at the points $A$ and $B$.\\
(i) The mid-point of $A B$ is $M$. Show that the coordinates of $M$ are $\left( \frac { 1 } { 2 } , 7 \frac { 1 } { 2 } \right)$.\\
(ii) Find the coordinates of the point $Q$ on the curve at which the tangent is parallel to the line $y + 3 x = 9$.\\
(iii) Find the distance $M Q$.
\hfill \mbox{\textit{CAIE P1 2004 Q5 [8]}}