OCR MEI Paper 1 2019 June — Question 2 3 marks

Exam BoardOCR MEI
ModulePaper 1 (Paper 1)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeIntersection of two lines
DifficultyModerate -0.8 This is a straightforward coordinate geometry question requiring students to find the gradient of the line through two points, write its equation, and show the lines are parallel (same gradient). It's simpler than average as it only requires basic techniques with no problem-solving insight, though the 'show that' format requires clear working.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships

2 Show that the line which passes through the points \(( 2 , - 4 )\) and \(( - 1,5 )\) does not intersect the line \(3 x + y = 10\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
(Gradient of line segment) \(= \frac{5-(-4)}{-1-2} = -3\)M1 AO3.1a — Attempt to find gradient; accept sign errors but not reciprocal
Given line \(y = -3x + 10\) has gradient \(-3\); same gradient so parallel linesM1 AO1.1a — Finding gradient of given line
Neither point lies on the line so the lines do not intersectA1 AO2.2a — Must conclude based on lines being parallel and not the same line
[3]
Alternative solution:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Gradient \(= \frac{5-(-4)}{-1-2} = -3\)M1 Attempt to find gradient; accept sign errors but not reciprocal
Equation of line: \(y-(-4) = -3(x-2)\), giving \(y = -3x+2\)M1 Finding equation of given line
Given line is \(y = -3x+10\) which is parallel, so lines do not intersectA1 Conclusion referring to parallel lines. Allow for solving two lines simultaneously and stating there are no solutions.
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| (Gradient of line segment) $= \frac{5-(-4)}{-1-2} = -3$ | M1 | AO3.1a — Attempt to find gradient; accept sign errors but not reciprocal |
| Given line $y = -3x + 10$ has gradient $-3$; same gradient so parallel lines | M1 | AO1.1a — Finding gradient of given line |
| Neither point lies on the line so the lines do not intersect | A1 | AO2.2a — Must conclude based on lines being parallel and not the same line |
| **[3]** | | |

**Alternative solution:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient $= \frac{5-(-4)}{-1-2} = -3$ | M1 | Attempt to find gradient; accept sign errors but not reciprocal |
| Equation of line: $y-(-4) = -3(x-2)$, giving $y = -3x+2$ | M1 | Finding equation of given line |
| Given line is $y = -3x+10$ which is parallel, so lines do not intersect | A1 | Conclusion referring to parallel lines. Allow for solving two lines simultaneously and stating there are no solutions. |
2 Show that the line which passes through the points $( 2 , - 4 )$ and $( - 1,5 )$ does not intersect the line $3 x + y = 10$.

\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q2 [3]}}