Edexcel C1 — Question 6 6 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeMidpoint of line segment
DifficultyModerate -0.5 This question requires finding axis intercepts, calculating a midpoint, then finding a distance—all standard C1 techniques. While it involves multiple steps and requires expressing the answer in surd form, these are routine procedures with no conceptual challenges, making it slightly easier than average.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.10f Distance between points: using position vectors
6. The straight line \(l\) has the equation \(x - 2 y = 12\) and meets the coordinate axes at the points \(A\) and \(B\). Find the distance of the mid-point of \(A B\) from the origin, giving your answer in the form \(k \sqrt { 5 }\).
AnswerMarks Guidance
\(x = 0 \Rightarrow y = -6\) ⇒ \((0, -6)\)B1
\(y = 0 \Rightarrow x = 12\) ⇒ \((12, 0)\)
mid-point \(= \left(\frac{0+12}{2}, \frac{-6+0}{2}\right) = (6, -3)\)M1 A1
dist. from \(O = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}\)M1, M1 A1 (6 marks) 
$x = 0 \Rightarrow y = -6$ ⇒ $(0, -6)$ | B1 |
$y = 0 \Rightarrow x = 12$ ⇒ $(12, 0)$ | |
mid-point $= \left(\frac{0+12}{2}, \frac{-6+0}{2}\right) = (6, -3)$ | M1 A1 |
dist. from $O = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$ | M1, M1 A1 | (6 marks)
6. The straight line $l$ has the equation $x - 2 y = 12$ and meets the coordinate axes at the points $A$ and $B$.

Find the distance of the mid-point of $A B$ from the origin, giving your answer in the form $k \sqrt { 5 }$.\\

\hfill \mbox{\textit{Edexcel C1  Q6 [6]}}
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