Home / iGCSE Mathematics (0580) – C3.2 Drawing linear graphs- Exam Style Questions Paper 3

iGCSE Mathematics (0580) - C3.2 Drawing linear graphs- Exam Style Questions Paper 3- New Syllabus

IGCSE Mathematics(0580) Practice Questions Paper 3 - All Chapters

Question

(a) Write down the coordinates of point P.

(b) On the grid, draw the line $y = x$.

(c) On the grid, draw the line that goes through point P and is perpendicular to the line $y = x$.

Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):

• C3.2 Drawing linear graphs
▶️ Answer/Explanation

(a)
Looking at the grid, point P is aligned with $-2$ on the x-axis and 3 on the y-axis.
✅ Answer: (-2, 3)

(b)
The line $y = x$ is a straight diagonal line passing through the origin (0,0), (1,1), (2,2), etc.
✅ Answer: Correct ruled line drawn passing through origin at 45 degrees.

(c)
A line perpendicular to $y = x$ will have a gradient of $-1$ (the negative reciprocal). Passing through P (-2, 3), the equation is $y = -x + 1$, which goes through (0, 1), (-1, 2), (-2, 3).

✅ Answer: Correct ruled line drawn for $y = -x + 1$ passing through P.

Question

(a) The table shows some values of \(y = \frac{6}{x}.\)

x-5-4-3-2-112345
y-1.2-1.5   6 21.51.2

(i) Complete the table.

(ii) On the grid, draw the graph of \(y = \frac{6}{x}\) for -5 ≤ x ≤ -1 and 1 ≤ x ≤ 5.

(iii) On the same grid, draw the line y = 4.

(iv) Find the co-ordinates of the point where the line y = 4 crosses the graph of \(y = \frac{6}{x}\).

(b)

(i) On this grid, plot the point A (–1, –3).

(ii) Draw a line with gradient 2 through point A.

(iii) Write down the equation of your line in the form y = mx + c.

▶️ Answer/Explanation
Solution

(a)(i) Ans:

x-5-4-3-2-112345
y-1.2-1.5-2-3-66321.51.2

Compute \(y = \frac{6}{x}\) for missing \(x\) values. For example, when \(x = -3\), \(y = \frac{6}{-3} = -2\). Similarly, fill all gaps.

(ii) Plot the points and draw two smooth hyperbola curves (one in the negative \(x\) region and one in the positive \(x\) region).

(iii) Draw a horizontal line at \(y = 4\).

(iv) Ans: (1.4 to 1.6, 4)

Solve \(\frac{6}{x} = 4\) to get \(x = 1.5\). Thus, the intersection point is \((1.5, 4)\).

(b)(i) Plot the point A at \((-1, -3)\).

(ii) Draw a line through A with slope 2 (i.e., for every 1 unit right, go 2 units up).

(iii) Ans: \(y = 2x – 1\)

Using point-slope form: \(y + 3 = 2(x + 1)\) simplifies to \(y = 2x – 1\).

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