Home / iGCSE Mathematics (0580) – C3.5 Equations of linear graphs- Exam Style Questions Paper 1

iGCSE Mathematics (0580) - C3.5 Equations of linear graphs- Exam Style Questions Paper 1- New Syllabus

Question

Find the equation of line L. Give your answer in the form y = mx + c.

▶️ Answer/Explanation
Solution

Ans: y = 3x + 2

From the graph, y-intercept (c) = 2.

Slope (m) = rise/run = (5-2)/(1-0) = 3/1 = 3.

Thus equation is y = 3x + 2.

Question

(a) Find the equation of line L in the form y = mx + c

(b) On the grid, draw a line that is perpendicular to line L

▶️ Answer/Explanation
Solution

Ans:

(a) y = (1/3)x + 1 (from the graph, slope is 1/3 and y-intercept is 1)

(b) Any line with slope -3 (negative reciprocal of 1/3) would be perpendicular

Question

Line $L$ is shown on the grid.

(a) Find the equation of line $L$ in the form $y = mx + c$.

(b) Line $L$ crosses the $x$-axis at $P$. Find the coordinates of $P$.

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

C3.3 Gradient of linear graphs
▶️ Answer/Explanation
(a) First, find where the line crosses the $y$-axis; it intersects at $(0, 3)$, so $c = 3$. Next, determine the gradient ($m$) by finding the rise over run between two points, such as $(0, 3)$ and $(2, 4)$, which gives $m = \frac{4 – 3}{2 – 0} = \frac{1}{2}$. Thus, the equation is $y = \frac{1}{2}x + 3$.
(b) Where the line hits the $x$-axis, the value of $y$ must be 0. Substituting $y = 0$ into our equation gives $0 = \frac{1}{2}x + 3$, which simplifies to $\frac{1}{2}x = -3$, meaning $x = -6$. So, the coordinates are $(-6, 0)$.
Answer:
(a) $y = \frac{1}{2}x + 3$
(b) (-6, 0)
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