| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry question requiring standard techniques: verifying a line equation using two points (part a), finding a midpoint, finding intersection with x-axis, then using distance formula and cosine rule or gradient methods to find an angle. All steps are routine AS-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| gradient \(AC = \frac{4-(-1)}{1-(-4)} = 1\) | M1 | Gradient calculation must be seen; o.e. using \(C(-4,-1)\); Also allow using \(y = x + c\) and evaluating \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Equation is \(y = x + 3\) | A1 | AG Must be clearly shown |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x=-4\), \(y = -4+3 = -1\) so C lies on the line | M1 | Checking both points lie on the line |
| Equation of line AC is \(y = x+3\) | A1 | Clear conclusion must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(M\) is \((4, 2)\) | B1 | |
| \(y = x+3\) crosses the \(x\)-axis at \(D(-3, 0)\) | B1 | |
| Length \(DM = \sqrt{49+4} = \sqrt{53}\) | M1 | Attempt to find at least two sides of triangle DMA; Also allow for \(AC^2\) etc found if clear |
| Answer | Marks | Guidance |
|---|---|---|
| Length \(DA = \sqrt{16+16} = \sqrt{32}\) | A1 | At least 2 correct lengths soi; FT their coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos D\hat{M}A = \frac{13+53-32}{2 \times \sqrt{13} \times \sqrt{53}}\) | M1 A1 | Allow sign errors, any form; FT their lengths. Must be correct expression for \(\cos D\hat{M}A\); Allow M1A0 for cosine rule leading to one of the other two angles \((101.3°\) or \(29.1°)\) |
| \(D\hat{M}A = 49.6°\) | A1 | Accept \(49.7°\) |
# Question 10:
## Part (a):
gradient $AC = \frac{4-(-1)}{1-(-4)} = 1$ | **M1** | Gradient calculation must be seen; o.e. using $C(-4,-1)$; Also allow using $y = x + c$ and evaluating $c$
$y - 4 = 1(x-1)$
Equation is $y = x + 3$ | **A1** | AG Must be clearly shown
**Alternative method:**
When $x=1$, $y = 1+3 = 4$ so A lies on the line
When $x=-4$, $y = -4+3 = -1$ so C lies on the line | **M1** | Checking both points lie on the line
Equation of line AC is $y = x+3$ | **A1** | Clear conclusion must be seen
## Part (b):
$M$ is $(4, 2)$ | **B1** |
$y = x+3$ crosses the $x$-axis at $D(-3, 0)$ | **B1** |
Length $DM = \sqrt{49+4} = \sqrt{53}$ | **M1** | Attempt to find at least two sides of triangle DMA; Also allow for $AC^2$ etc found if clear
Length $MA = \sqrt{9+4} = \sqrt{13}$
Length $DA = \sqrt{16+16} = \sqrt{32}$ | **A1** | At least 2 correct lengths soi; FT their coordinates
Using the cosine rule:
$\cos D\hat{M}A = \frac{13+53-32}{2 \times \sqrt{13} \times \sqrt{53}}$ | **M1 A1** | Allow sign errors, any form; FT their lengths. Must be correct expression for $\cos D\hat{M}A$; Allow M1A0 for cosine rule leading to one of the other two angles $(101.3°$ or $29.1°)$
$D\hat{M}A = 49.6°$ | **A1** | Accept $49.7°$
---10 A triangle has vertices $A ( 1,4 ) , B ( 7,0 )$ and $C ( - 4 , - 1 )$.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the line AC is $\mathrm { y } = \mathrm { x } + 3$.
M is the midpoint of AB . The line AC intersects the $x$-axis at D .
\item Determine the angle DMA.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2022 Q10 [9]}}