iGCSE Maths (0580) C 3.2 Drawing linear graphs Paper 1- Exam Style Questions- New Syllabus
Question
(a) Line A has equation $y = 3x + 1$. Line B has equation $y = 3x – 1$.
Draw a ring around the description that is correct:
• Line A intersects line B
• Line A has a steeper gradient than line B
• Line A is perpendicular to line B
• Line A is parallel to line B
• Line A and Line B intersect the y-axis at the same point
(b)

On the grid, draw the graph of $y = 2x – 1$.
Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):
▶️ Answer/Explanation
(a) Both linear equations are written in the standard form $y = mx + c$, where $m$ is the gradient. For line A, the gradient is 3, and for line B, the gradient is also 3. Since they have identical gradients but different y-intercepts ($+1$ and $-1$), the lines run in exactly the same direction and will never meet. Thus, Line A is parallel to line B.
(b) To plot $y = 2x – 1$, we can find a couple of coordinates:
• If $x = 0$, $y = 2(0) – 1 = -1 \rightarrow (0, -1)$
• If $x = 1$, $y = 2(1) – 1 = 1 \rightarrow (1, 1)$
• If $x = 2$, $y = 2(2) – 1 = 3 \rightarrow (2, 3)$ Plot these points on the provided coordinate grid and draw a single straight line through them using a ruler.
Answers:
(a) Ring around: Line A is parallel to line B
(b) [A straight line passing through $(0, -1)$, $(1, 1)$, and $(2, 3)$]

Mark the midpoint of the line ST.
▶️ Answer/Explanation
Ans: Midpoint of ST marked
To find the midpoint of line segment ST:
- Identify the coordinates of points S and T from the diagram.
- Use the midpoint formula: \[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] where \((x_1, y_1)\) are the coordinates of S and \((x_2, y_2)\) are the coordinates of T.
- Mark the calculated midpoint on the line ST.
