| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Verify collinearity or parallel/perpendicular relationship |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding midpoint, gradients, perpendicularity check, and tan formula. Part (a) is routine calculation (6 marks for showing perpendicular gradients multiply to -1), part (b) uses standard tan formula with two gradients, and part (c) is recognition. All techniques are basic A-level content with no problem-solving insight required, making it easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) A correct method for finding the coordinates of the mid-point of \(AB\) | M1 | AO1 |
| \(D\) has coordinates \((-1, 5)\) | A1 | AO1 |
| Gradient of \(AB = \frac{\text{increase in } y}{\text{increase in } x}\) | M1 | AO1 |
| Gradient of \(AB = \frac{6}{2}\) | A1 | AO1 (or equivalent) |
| Gradient of \(CD = \frac{\text{increase in } y}{\text{increase in } x}\) | (M1) | (AO1) (to be awarded only if the previous M1 is not awarded) |
| Gradient of \(CD = \frac{7}{21}\) | A1 | AO1 (or equivalent) |
| \(-\frac{6}{2} \times \frac{7}{21} = -1 \Rightarrow AB\) is perpendicular to \(CD\) | B1 | AO2 |
| (b) A correct method for finding the length of \(AD\) or \(CD\) | M1 | AO1 |
| \(AD = \sqrt{10}\) | A1 | AO1 |
| \(CD = \sqrt{490}\) | A1 | AO1 |
| \(\tan \angle CAB = \frac{CD}{AD}\) | M1 | AO1 |
| \(\tan \angle CAB = 7\) | A1 | AO1 |
| (c) Isosceles | B1 | AO2 |
**(a)** A correct method for finding the coordinates of the mid-point of $AB$ | M1 | AO1
$D$ has coordinates $(-1, 5)$ | A1 | AO1
Gradient of $AB = \frac{\text{increase in } y}{\text{increase in } x}$ | M1 | AO1
Gradient of $AB = \frac{6}{2}$ | A1 | AO1 (or equivalent)
Gradient of $CD = \frac{\text{increase in } y}{\text{increase in } x}$ | (M1) | (AO1) (to be awarded only if the previous M1 is not awarded)
Gradient of $CD = \frac{7}{21}$ | A1 | AO1 (or equivalent)
$-\frac{6}{2} \times \frac{7}{21} = -1 \Rightarrow AB$ is perpendicular to $CD$ | B1 | AO2
**(b)** A correct method for finding the length of $AD$ or $CD$ | M1 | AO1
$AD = \sqrt{10}$ | A1 | AO1
$CD = \sqrt{490}$ | A1 | AO1
$\tan \angle CAB = \frac{CD}{AD}$ | M1 | AO1
$\tan \angle CAB = 7$ | A1 | AO1
**(c)** Isosceles | B1 | AO2
**Total: [12]**
---The points $A(0, 2)$, $B(-2, 8)$, $C(20, 12)$ are the vertices of the triangle $ABC$. The point $D$ is the mid-point of $AB$.
\begin{enumerate}[label=(\alph*)]
\item Show that $CD$ is perpendicular to $AB$. [6]
\item Find the exact value of $\tan CAB$. [5]
\item Write down the geometrical name for the triangle $ABC$. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 1 Q5 [12]}}