CAIE P1 2011 June — Question 3 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2011
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeCoordinates from geometric constraints
DifficultyStandard +0.3 This is a straightforward coordinate geometry problem requiring students to find intercepts (P = (a,0), Q = (0,b)), apply the distance formula (a² + b² = 45), use the gradient formula (-b/a = -1/2), then solve the resulting simultaneous equations. It involves standard techniques with no novel insight required, making it slightly easier than average.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.10c Magnitude and direction: of vectors
3 The line \(\frac { x } { a } + \frac { y } { b } = 1\), where \(a\) and \(b\) are positive constants, meets the \(x\)-axis at \(P\) and the \(y\)-axis at \(Q\). Given that \(P Q = \sqrt { } ( 45 )\) and that the gradient of the line \(P Q\) is \(- \frac { 1 } { 2 }\), find the values of \(a\) and \(b\).
\(\frac{x}{a} + \frac{y}{b} = 1\)
\(P(a, 0)\) and \(Q(0, b)\)
AnswerMarks Guidance
Distance \(\rightarrow \sqrt{(a^2 + b^2)} = \sqrt{45}\)M1 A1
Gradients \(\rightarrow \frac{-a}{b} = \frac{-1}{2}\)M1 A1 M1 even if sign(s) incorrect.
Solution of sim eqns \(\rightarrow a = 6, b = 3\)A1 Correct values \(a\) and \(b\) (both)
[5] 
$\frac{x}{a} + \frac{y}{b} = 1$

$P(a, 0)$ and $Q(0, b)$

Distance $\rightarrow \sqrt{(a^2 + b^2)} = \sqrt{45}$ | M1 A1 |
Gradients $\rightarrow \frac{-a}{b} = \frac{-1}{2}$ | M1 A1 | M1 even if sign(s) incorrect.
Solution of sim eqns $\rightarrow a = 6, b = 3$ | A1 | Correct values $a$ and $b$ (both)
| [5] |
3 The line $\frac { x } { a } + \frac { y } { b } = 1$, where $a$ and $b$ are positive constants, meets the $x$-axis at $P$ and the $y$-axis at $Q$. Given that $P Q = \sqrt { } ( 45 )$ and that the gradient of the line $P Q$ is $- \frac { 1 } { 2 }$, find the values of $a$ and $b$.

\hfill \mbox{\textit{CAIE P1 2011 Q3 [5]}}
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