SHS Core Mathematics · Year 1

Linear Equations, Relations and Functions practice questions

Change of subject of a formula, solving linear equations and inequalities in one and two variables, relations and functions (including mappings), and the gradient, equation and distance between two points on a straight line.

26 questions in this topic: 8 recall, 13 applying skills and concepts, 4 strategic thinking and 1 extended thinking. Every one carries a written explanation.

Sample questions

5 of 26 questions in this topic. Sign up to answer them and see the explanations.

Question 1Strategic thinking
Using the point-gradient formula y − y₁ = m(x − x₁), the equation of the line with gradient −2 that passes through the point (3, −4) is
- ay = −2x + 2
- by = −2x − 2
- cy = 2x − 4
- dy = −2x + 4
Question 2Skills & concepts
A line has gradient 2. The gradient of a line that is perpendicular to it is
- a−1/2
- b2
- c1/2
- d−2
Question 3Skills & concepts
The gradient of the straight line joining P(−4, 5) and Q(4, 17) is
- a1.5
- b12
- c8
- d0.5
Question 4Skills & concepts
Rewriting the equation 3x − 7y = 16 into the form y = mx + c to identify its gradient gives y = (3/7)x − 16/7, so this equation
- arepresents a straight line, with gradient 3/7
- bdoes not represent a straight line at all
- crepresents a straight line with gradient 16/7
- dcannot be rearranged into the form y = mx + c
Question 5Recall
The distance (magnitude) between two points P(x₁, y₁) and Q(x₂, y₂), derived from the Pythagorean theorem, is given by
- a|PQ| = √((x₂ − x₁)² + (y₂ − y₁)²)
- b|PQ| = (x₂ − x₁) + (y₂ − y₁)
- c|PQ| = (x₂ − x₁)² + (y₂ − y₁)²
- d|PQ| = √(x₂ + x₁) + √(y₂ + y₁)

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Linear Equations, Relations and Functions practice questions

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