CAIE P1 2005 November — Question 7 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2005
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypePerpendicular bisector of segment
DifficultyModerate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding midpoint, gradient of AB, perpendicular gradient, equation of perpendicular bisector, equation of BC, then solving simultaneous equations. All steps are routine textbook procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships

7 Three points have coordinates \(A ( 2,6 ) , B ( 8,10 )\) and \(C ( 6,0 )\). The perpendicular bisector of \(A B\) meets the line \(B C\) at \(D\). Find
  1. the equation of the perpendicular bisector of \(A B\) in the form \(a x + b y = c\),
  2. the coordinates of \(D\).

(i) \(M(5,8)\)
AnswerMarks Guidance
gradient of \(AB = \frac{2}{3}\), Perp= \(-\frac{3}{2}\)B1 M1 Co. Use of step/x-step + \(m_1m_2=-1\) for AB
Equation \(y - 8 = -\frac{3}{2}(x-5)\)M1 Use of \(y-k=m(x-h)\) not \((y+k)\) etc
\(\rightarrow 2y + 3x = 31\) (or locus method M1A1M1A1)A1 [4]
(ii) \(BC. y = 5(x-6) \rightarrow y = 5x - 30\)
AnswerMarks Guidance
Sim Eqns \(\rightarrow (7,5)\)M1A1 DM1A1 Use of \(y-k=m(x-h)\) not \((y+k)\) etc. co Correct attempt at soln of \(BC\) with his answer to (i).
(i) $M(5,8)$
gradient of $AB = \frac{2}{3}$, Perp= $-\frac{3}{2}$ | B1 M1 | Co. Use of step/x-step + $m_1m_2=-1$ for AB

Equation $y - 8 = -\frac{3}{2}(x-5)$ | M1 | Use of $y-k=m(x-h)$ not $(y+k)$ etc

$\rightarrow 2y + 3x = 31$ (or locus method M1A1M1A1) | A1 | [4]

(ii) $BC. y = 5(x-6) \rightarrow y = 5x - 30$
Sim Eqns $\rightarrow (7,5)$ | M1A1 DM1A1 | Use of $y-k=m(x-h)$ not $(y+k)$ etc. co Correct attempt at soln of $BC$ with his answer to (i). | [4]
7 Three points have coordinates $A ( 2,6 ) , B ( 8,10 )$ and $C ( 6,0 )$. The perpendicular bisector of $A B$ meets the line $B C$ at $D$. Find\\
(i) the equation of the perpendicular bisector of $A B$ in the form $a x + b y = c$,\\
(ii) the coordinates of $D$.

\hfill \mbox{\textit{CAIE P1 2005 Q7 [8]}}