CAIE P1 2013 June — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeSymmetric or reflected points
DifficultyChallenging +1.2 This requires multiple coordinated steps: finding the perpendicular from (-1,3) to the line, solving simultaneously for the foot of perpendicular, then using midpoint properties to find the reflection. While each individual step uses standard techniques (perpendicular gradients, simultaneous equations, midpoint formula), the multi-step coordination and spatial reasoning elevate this above routine exercises.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships

7 The point \(R\) is the reflection of the point \(( - 1,3 )\) in the line \(3 y + 2 x = 33\). Find by calculation the coordinates of \(R\).

\(3y + 2x = 33\)
Gradient of line \(= -\frac{2}{3}\)
Gradient of perpendicular \(= \frac{3}{2}\)
AnswerMarks Guidance
Eqn of perp: \(y - 3 = \frac{3}{2}(x + 1)\)B1 M1 M1 M1 A1 Use of \(m_1m_2 = -1\) with gradient of line. Correct form of perpendicular eqn. Sim eqns
Sim Eqns → \((3, 9)\)
AnswerMarks Guidance
\((-1, 3) \rightarrow (3, 9) \rightarrow (7, 15)\)M1 A1 Vectors or other method
[7]
$3y + 2x = 33$

Gradient of line $= -\frac{2}{3}$

Gradient of perpendicular $= \frac{3}{2}$

Eqn of perp: $y - 3 = \frac{3}{2}(x + 1)$ | B1 M1 M1 M1 A1 | Use of $m_1m_2 = -1$ with gradient of line. Correct form of perpendicular eqn. Sim eqns

Sim Eqns → $(3, 9)$

$(-1, 3) \rightarrow (3, 9) \rightarrow (7, 15)$ | M1 A1 | Vectors or other method

[7] |

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7 The point $R$ is the reflection of the point $( - 1,3 )$ in the line $3 y + 2 x = 33$. Find by calculation the coordinates of $R$.

\hfill \mbox{\textit{CAIE P1 2013 Q7 [7]}}