CAIE P1 2012 June — Question 4 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeDistance between two points
DifficultyModerate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding the midpoint of AB, calculating the gradient of AB, determining the perpendicular gradient, writing the equation of the perpendicular bisector, finding intercepts, and calculating distance. All steps are routine applications of formulas with no problem-solving insight required, 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

The point \(A\) has coordinates \((-1, -5)\) and the point \(B\) has coordinates \((7, 1)\). The perpendicular bisector of \(AB\) meets the \(x\)-axis at \(C\) and the \(y\)-axis at \(D\). Calculate the length of \(CD\). [6]

AnswerMarks Guidance
\(A(-1, -5)\), \(B(7, 1)\), \(M(3, -2)\)B1 co
Gradient = \(\frac{3}{4}\)B1 co
Perpendicular gradient = \(-\frac{4}{3}\)B1 co needs to be perp. through M. Setting one of x or y to 0.
Eqn \(y + 2 = -\frac{4}{3}(x - 3)\)M1 Correct method. co.
Sets x and y to 0 \(C(\frac{3}{4}, 0)\), \(D(0, 2)\)M1
\(\to\) Pythagoras \(\to\) \(CD = 2.5\)M1 A1 [6]
$A(-1, -5)$, $B(7, 1)$, $M(3, -2)$ | B1 | co

Gradient = $\frac{3}{4}$ | B1 | co

Perpendicular gradient = $-\frac{4}{3}$ | B1 | co needs to be perp. through M. Setting one of x or y to 0.

Eqn $y + 2 = -\frac{4}{3}(x - 3)$ | M1 | Correct method. co.

Sets x and y to 0 $C(\frac{3}{4}, 0)$, $D(0, 2)$ | M1 | 

$\to$ Pythagoras $\to$ $CD = 2.5$ | M1 A1 | [6]
The point $A$ has coordinates $(-1, -5)$ and the point $B$ has coordinates $(7, 1)$. The perpendicular bisector of $AB$ meets the $x$-axis at $C$ and the $y$-axis at $D$. Calculate the length of $CD$. [6]

\hfill \mbox{\textit{CAIE P1 2012 Q4 [6]}}