| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Gradient from equation or points |
| Difficulty | Easy -1.3 This is a straightforward C1 question testing basic rearrangement of linear equations to find gradient and intercepts, followed by a simple midpoint calculation. Both parts require only routine algebraic manipulation with no problem-solving insight needed, making it easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3}{5}\) | B1 [1] | Allow \(0.6\) or any equivalent fraction. Do not allow \(\frac{3}{5}x\) as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\left(\frac{20}{3}, 0\right)\) | B1 | May be implied by subsequent working. Allow \(x = \frac{20}{3}\) for P |
| \(Q(0, -4)\) | B1 | May be implied. Allow \(y = -4\) for Q |
| \(\left(\frac{\frac{20}{3}+0}{2}, \frac{0+-4}{2}\right)\) | M1 | Correct method to find midpoint of line. Check formula, or if formula not seen, the use of formula is correct (including correct signs) for both \(x\) and \(y\). Can be implied by correct final answers. SC |
| \(\left(\frac{10}{3}, -2\right)\) | A1 [4] | Allow exact equivalent forms, decimals must be correct to at least 2dp. If P and Q given the wrong way round but then used correctly to obtain correct final answer B2 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3}{5}$ | B1 [1] | Allow $0.6$ or any equivalent fraction. Do not allow $\frac{3}{5}x$ as final answer |
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## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\left(\frac{20}{3}, 0\right)$ | B1 | May be implied by subsequent working. Allow $x = \frac{20}{3}$ for P |
| $Q(0, -4)$ | B1 | May be implied. Allow $y = -4$ for Q |
| $\left(\frac{\frac{20}{3}+0}{2}, \frac{0+-4}{2}\right)$ | M1 | Correct method to find midpoint of line. Check formula, or if formula not seen, the use of formula is correct (including correct signs) for both $x$ and $y$. Can be implied by correct final answers. SC |
| $\left(\frac{10}{3}, -2\right)$ | A1 [4] | Allow exact equivalent forms, decimals must be correct to at least 2dp. If P and Q given the wrong way round but then used correctly to obtain correct final answer **B2** |
---3 (i) Find the gradient of the line $l$ which has equation $3 x - 5 y - 20 = 0$.\\
(ii) The line $l$ crosses the $x$-axis at $P$ and the $y$-axis at $Q$. Find the coordinates of the mid-point of $P Q$.
\hfill \mbox{\textit{OCR C1 2012 Q3 [5]}}