AQA C1 2015 June — Question 2 5 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeGradient from equation or points
DifficultyModerate -0.8 This is a straightforward gradient calculation using the formula (y₂-y₁)/(x₂-x₁), followed by rationalizing the denominator—a standard algebraic manipulation. While the surds add minor computational complexity beyond a basic gradient question, this remains a routine C1 exercise requiring only recall of standard techniques with no problem-solving insight needed.
Spec1.02b Surds: manipulation and rationalising denominators1.03a Straight lines: equation forms y=mx+c, ax+by+c=0

2 The point \(P\) has coordinates \(( \sqrt { 3 } , 2 \sqrt { 3 } )\) and the point \(Q\) has coordinates \(( \sqrt { 5 } , 4 \sqrt { 5 } )\). Show that the gradient of \(P Q\) can be expressed as \(n + \sqrt { 15 }\), stating the value of the integer \(n\).
[0pt] [5 marks]

Question 2:
AnswerMarks Guidance
Gradient \(= \frac{4\sqrt{5} - 2\sqrt{3}}{\sqrt{5} - \sqrt{3}}\)M1 Correct gradient formula applied
Multiply numerator and denominator by \((\sqrt{5} + \sqrt{3})\)M1 Rationalising the denominator
Numerator: \(4\sqrt{5}(\sqrt{5}+\sqrt{3}) - 2\sqrt{3}(\sqrt{5}+\sqrt{3}) = 20 + 4\sqrt{15} - 2\sqrt{15} - 6 = 14 + 2\sqrt{15}\)A1 Correct numerator expansion
Denominator: \((\sqrt{5})^2 - (\sqrt{3})^2 = 2\)A1 Correct denominator
Gradient \(= \frac{14 + 2\sqrt{15}}{2} = 7 + \sqrt{15}\), so \(n = 7\)A1 Correct final answer with \(n = 7\) stated
# Question 2:

| Gradient $= \frac{4\sqrt{5} - 2\sqrt{3}}{\sqrt{5} - \sqrt{3}}$ | M1 | Correct gradient formula applied |
| Multiply numerator and denominator by $(\sqrt{5} + \sqrt{3})$ | M1 | Rationalising the denominator |
| Numerator: $4\sqrt{5}(\sqrt{5}+\sqrt{3}) - 2\sqrt{3}(\sqrt{5}+\sqrt{3}) = 20 + 4\sqrt{15} - 2\sqrt{15} - 6 = 14 + 2\sqrt{15}$ | A1 | Correct numerator expansion |
| Denominator: $(\sqrt{5})^2 - (\sqrt{3})^2 = 2$ | A1 | Correct denominator |
| Gradient $= \frac{14 + 2\sqrt{15}}{2} = 7 + \sqrt{15}$, so $n = 7$ | A1 | Correct final answer with $n = 7$ stated |

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2 The point $P$ has coordinates $( \sqrt { 3 } , 2 \sqrt { 3 } )$ and the point $Q$ has coordinates $( \sqrt { 5 } , 4 \sqrt { 5 } )$. Show that the gradient of $P Q$ can be expressed as $n + \sqrt { 15 }$, stating the value of the integer $n$.\\[0pt]
[5 marks]

\hfill \mbox{\textit{AQA C1 2015 Q2 [5]}}