| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Easy -1.3 This is a straightforward textbook exercise requiring only the standard two-step process: calculate gradient m = (y₂-y₁)/(x₂-x₁), then substitute into y = mx + c to find c. It's routine recall with minimal problem-solving, easier than average A-level questions. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| VI4V SIHI NI JIIIM ION OC | VI4V SIHI NI JALYM ION OC | VJYV SIHI NI JLIYM ION OO |
| VIIIV SIHI NI JIIIM ION OC | VIAV SIHI NI JALM IONOO | VJYV SIHL NI GLIYM LON OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Gradient \(= -3\) | B1 | For the correct gradient. If candidate starts \(\frac{10--2}{1-5} = \frac{y-10}{x-1}\) then not awarded until \(-3\) is seen |
| Uses their gradient and a point e.g. \(y + 2 = -3(x-5)\) | M1 | For a correct attempt at finding equation of \(l\) using their numerical gradient and any point on the line. Mid-point \((3,4)\) may be used. Allow if candidate makes a slip on gradient and/or a sign on a coordinate. e.g. \(m = \frac{10-2}{1-5} = -2\) then \(y-2=-2(x-5)\). If candidate finds \(m=-3\) then attempts perpendicular line it is M0 |
| \(\Rightarrow y = -3x + 13\) | A1 | CAO \(y = -3x+13\) or \(y = 13-3x\). Do not allow leaving as \(m=-3, c=13\) unless preceded by correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Reaches \(m = -3\) | B1 | |
| Attempts to solve simultaneously to get \(m = \ldots\) and \(c = \ldots\) | M1 | Do not look into mechanics of the attempt; sight of values for \(m\) and \(c\) are sufficient |
| \(y = -3x + 13\) | A1 | Do not allow leaving as \(m=-3, c=13\) |
## Question 1:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Gradient $= -3$ | B1 | For the correct gradient. If candidate starts $\frac{10--2}{1-5} = \frac{y-10}{x-1}$ then not awarded until $-3$ is seen |
| Uses **their** gradient and a point e.g. $y + 2 = -3(x-5)$ | M1 | For a correct attempt at finding equation of $l$ using **their** numerical gradient and any point on the line. Mid-point $(3,4)$ may be used. Allow if candidate makes a slip on gradient and/or a sign on a coordinate. e.g. $m = \frac{10-2}{1-5} = -2$ then $y-2=-2(x-5)$. If candidate finds $m=-3$ then attempts perpendicular line it is M0 |
| $\Rightarrow y = -3x + 13$ | A1 | CAO $y = -3x+13$ or $y = 13-3x$. Do not allow leaving as $m=-3, c=13$ unless preceded by correct form |
**Alternative (simultaneous equations approach):**
Using $-2 = 5m+c$ and $10 = 1m+c$, with attempt to place $(x,y)$ coordinates correctly:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Reaches $m = -3$ | B1 | |
| Attempts to solve simultaneously to get $m = \ldots$ and $c = \ldots$ | M1 | Do not look into mechanics of the attempt; sight of values for $m$ and $c$ are sufficient |
| $y = -3x + 13$ | A1 | Do not allow leaving as $m=-3, c=13$ |
**Total: (3 marks)**\begin{enumerate}
\item A line $l$ passes through the points $A ( 5 , - 2 )$ and $B ( 1,10 )$.
\end{enumerate}
Find the equation of $l$, writing your answer in the form $y = m x + c$ where $m$ and $c$ are constants.\\
(3)\\
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VI4V SIHI NI JIIIM ION OC & VI4V SIHI NI JALYM ION OC & VJYV SIHI NI JLIYM ION OO \\
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VIIIV SIHI NI JIIIM ION OC & VIAV SIHI NI JALM IONOO & VJYV SIHL NI GLIYM LON OO \\
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\hfill \mbox{\textit{Edexcel C12 2019 Q1 [3]}}