CAIE P1 2007 November — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2007
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeCoordinates from geometric constraints
DifficultyStandard +0.3 This is a coordinate geometry problem requiring calculation of gradients, perpendicular/parallel line conditions, and solving simultaneous equations. While it involves multiple steps (finding gradient of AB, using perpendicular condition for DA, using parallel condition for DC, finding intersection), these are all standard techniques taught early in P1. The problem is slightly above average difficulty due to the need to coordinate multiple conditions, but remains a straightforward application of core principles without requiring novel insight.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships

6 \includegraphics[max width=\textwidth, alt={}, center]{e753f588-97bc-4c6a-a82b-7b6a6d0cadc4-2_627_748_1685_699} The three points \(A ( 3,8 ) , B ( 6,2 )\) and \(C ( 10,2 )\) are shown in the diagram. The point \(D\) is such that the line \(D A\) is perpendicular to \(A B\) and \(D C\) is parallel to \(A B\). Calculate the coordinates of \(D\).

AnswerMarks Guidance
Gradient of \(AB = -2\); Eqn of \(CD\): \(y - 2 = -2(x - 10)\) (\(y + 2x = 22\))B1, M1 A1∨ co correct form of eqn (inc \(y = mx + c\)) – awarded for either \(CD\) or \(AD\). accept any form for A mark.
Uses \(m_1m_2 = -1\); Eqn of \(DA\): \(y - 8 = \frac{1}{2}(x - 3)\) (\(2y = x + 13\))M1, A1∨ [7] Use of \(m_1m_2 = -1\); Any correct form.
Sim eqns → \((6.2, 9.6)\)M1A1 [7] Reasonable algebra. co.
Gradient of $AB = -2$; Eqn of $CD$: $y - 2 = -2(x - 10)$ ($y + 2x = 22$) | B1, M1 A1∨ | co correct form of eqn (inc $y = mx + c$) – awarded for either $CD$ or $AD$. accept any form for A mark.
Uses $m_1m_2 = -1$; Eqn of $DA$: $y - 8 = \frac{1}{2}(x - 3)$ ($2y = x + 13$) | M1, A1∨ [7] | Use of $m_1m_2 = -1$; Any correct form.
Sim eqns → $(6.2, 9.6)$ | M1A1 [7] | Reasonable algebra. co.
6\\
\includegraphics[max width=\textwidth, alt={}, center]{e753f588-97bc-4c6a-a82b-7b6a6d0cadc4-2_627_748_1685_699}

The three points $A ( 3,8 ) , B ( 6,2 )$ and $C ( 10,2 )$ are shown in the diagram. The point $D$ is such that the line $D A$ is perpendicular to $A B$ and $D C$ is parallel to $A B$. Calculate the coordinates of $D$.

\hfill \mbox{\textit{CAIE P1 2007 Q6 [7]}}