| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Moderate -0.8 This is a straightforward multi-part coordinate geometry question testing standard techniques: finding line equations from point and gradient, solving simultaneous equations for intersection, finding midpoints, and verifying a line passes through a point. All parts are routine C1-level exercises requiring no problem-solving insight, though the multi-step nature and 11 total marks make it slightly more substantial than the most trivial questions. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = \frac{3}{2}x - \frac{9}{2}\) | M1 A1 | |
| (ii) \(3x - 4(\frac{1}{2}x - \frac{9}{2}) + 3 = 0\) | M1 | |
| \(x = 7 \therefore B(7, 6)\) | A2 | |
| (iii) \(= (\frac{2+7}{2}, \frac{1+6}{2}) = (6, \frac{9}{2})\) | M1 A1 | |
| (iv) \(l_2: y = \frac{2}{3}x + \frac{4}{3} \therefore \text{grad} = \frac{2}{3}\) | B1 | |
| \(\therefore y - \frac{2}{3} = \frac{4}{3}(x - 6)\) | M1 | |
| \(y = \frac{4}{3}x\) | A1 | |
| When \(x = 0, y = 0 \therefore\) passes through origin | A1 | (11) |
**(i)** $y = \frac{3}{2}x - \frac{9}{2}$ | M1 A1 |
**(ii)** $3x - 4(\frac{1}{2}x - \frac{9}{2}) + 3 = 0$ | M1 |
$x = 7 \therefore B(7, 6)$ | A2 |
**(iii)** $= (\frac{2+7}{2}, \frac{1+6}{2}) = (6, \frac{9}{2})$ | M1 A1 |
**(iv)** $l_2: y = \frac{2}{3}x + \frac{4}{3} \therefore \text{grad} = \frac{2}{3}$ | B1 |
$\therefore y - \frac{2}{3} = \frac{4}{3}(x - 6)$ | M1 |
$y = \frac{4}{3}x$ | A1 |
When $x = 0, y = 0 \therefore$ passes through origin | A1 | (11) |The straight line $l_1$ has gradient $\frac{3}{4}$ and passes through the point $A (5, 3)$.
\begin{enumerate}[label=(\roman*)]
\item Find an equation for $l_1$ in the form $y = mx + c$. [2]
\end{enumerate}
The straight line $l_2$ has the equation $3x - 4y + 3 = 0$ and intersects $l_1$ at the point $B$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the coordinates of $B$. [3]
\item Find the coordinates of the mid-point of $AB$. [2]
\item Show that the straight line parallel to $l_2$ which passes through the mid-point of $AB$ also passes through the origin. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q7 [11]}}