| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular from point to line |
| Difficulty | Standard +0.3 This question involves standard coordinate geometry techniques: calculating distances to verify isosceles property, finding area using the formula ½|base×height|, and finding the foot of perpendicular using gradient relationships and simultaneous equations. While multi-part with several steps, each component uses routine A-level methods without requiring novel insight or complex problem-solving. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.10a Vectors in 2D: i,j notation and column vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2 = 6^2 + 7^2 = 85\), \(BC^2 = 2^2 + 9^2 = 85\) \((\rightarrow \text{isosceles})\) | B1B1 | Or \(AB = BC = \sqrt{85}\) etc |
| \(AC^2 = 8^2 + 2^2 = 68\) | B1 | |
| \(M = (2, -2)\) or \(BM^2 = (\sqrt{85})^2 - (\frac{1}{2}\sqrt{68})^2\) | B1 | Where \(M\) is mid-point of \(AC\) |
| \(BM = \sqrt{2^2 + 8^2} = \sqrt{68}\) or \(\sqrt{85-17} = \sqrt{68}\) | B1 | |
| Area \(\triangle ABC = \dfrac{1}{2}\sqrt{68}\sqrt{68} = 34\) | B1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of \(AB = 7/6\) | B1 | |
| Equation of \(AB\) is \(y + 1 = \dfrac{7}{6}(x+2)\) | M1 | Or \(y - 6 = \dfrac{7}{6}(x-4)\) |
| Gradient of \(CD = -6/7\) | M1 | |
| Equation of \(CD\) is \(y + 3 = \dfrac{-6}{7}(x-6)\) | M1 | |
| Sim Eqns \(2 = \dfrac{-6}{7}x + \dfrac{36}{7} - \dfrac{7}{6}x - \dfrac{14}{6}\) | M1 | |
| \(x = \dfrac{34}{85} = \dfrac{2}{5}\) oe | A1 | |
| [6] |
## Question 11(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2 = 6^2 + 7^2 = 85$, $BC^2 = 2^2 + 9^2 = 85$ $(\rightarrow \text{isosceles})$ | B1B1 | Or $AB = BC = \sqrt{85}$ etc |
| $AC^2 = 8^2 + 2^2 = 68$ | B1 | |
| $M = (2, -2)$ or $BM^2 = (\sqrt{85})^2 - (\frac{1}{2}\sqrt{68})^2$ | B1 | Where $M$ is mid-point of $AC$ |
| $BM = \sqrt{2^2 + 8^2} = \sqrt{68}$ or $\sqrt{85-17} = \sqrt{68}$ | B1 | |
| Area $\triangle ABC = \dfrac{1}{2}\sqrt{68}\sqrt{68} = 34$ | B1 | |
| | [6] | |
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## Question 11(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $AB = 7/6$ | B1 | |
| Equation of $AB$ is $y + 1 = \dfrac{7}{6}(x+2)$ | M1 | Or $y - 6 = \dfrac{7}{6}(x-4)$ |
| Gradient of $CD = -6/7$ | M1 | |
| Equation of $CD$ is $y + 3 = \dfrac{-6}{7}(x-6)$ | M1 | |
| Sim Eqns $2 = \dfrac{-6}{7}x + \dfrac{36}{7} - \dfrac{7}{6}x - \dfrac{14}{6}$ | M1 | |
| $x = \dfrac{34}{85} = \dfrac{2}{5}$ oe | A1 | |
| | [6] | |11 Triangle $A B C$ has vertices at $A ( - 2 , - 1 ) , B ( 4,6 )$ and $C ( 6 , - 3 )$.\\
(i) Show that triangle $A B C$ is isosceles and find the exact area of this triangle.\\
(ii) The point $D$ is the point on $A B$ such that $C D$ is perpendicular to $A B$. Calculate the $x$-coordinate of $D$.
\hfill \mbox{\textit{CAIE P1 2016 Q11 [12]}}