Graphs & Coordinate Geometry

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Corbett MathsStraight line graphs — gradient and y-intercept

Corbett MathsParallel and perpendicular lines

Corbett MathsGraph transformations

GCSE Maths TutorGraph Transformations | Grade 7-9 Maths Series | GCSE Maths Tutor

GCSE Maths TutorCoordinate Geometry - Perpendicular Lines | Part 2 | Grade 7-9 Maths Series | GCSE Maths Tutor

GCSE Maths TutorEquations of Tangents to Circles (Grade 9) | Part 4 | Grade 9 Maths Series | GCSE Maths Tutor

Key facts & methods

Straight line graphs

y = mx + c: m = gradient, c = y-intercept.
Gradient = rise/run = (y₂−y₁)/(x₂−x₁).
Midpoint of (x₁,y₁) and (x₂,y₂) = ((x₁+x₂)/2, (y₁+y₂)/2).
Parallel lines: same gradient. If y=3x+1 is parallel to y=3x+c.
Perpendicular lines: gradients multiply to −1. If gradient is m, perpendicular gradient is −1/m.
E.g. Perpendicular to gradient 2/3 has gradient −3/2.

Graph transformations

y = f(x) is the original graph.
y = f(x) + a: translation up by a (down if a negative).
y = f(x + a): translation LEFT by a (right if a negative). Note: + means left.
y = af(x): stretch by scale factor a in y direction.
y = f(ax): stretch by scale factor 1/a in x direction.
y = −f(x): reflection in x-axis.
y = f(−x): reflection in y-axis.

Quadratic and other graphs

Quadratic y=ax²+bx+c: parabola. Roots = x-intercepts. Vertex found by completing the square.
Cubic y=ax³: S-shaped curve through origin.
Reciprocal y=1/x: two branches, asymptotes at x=0 and y=0.
Exponential y=aˣ: always positive, passes through (0,1), asymptote at y=0.
Circle (x−a)²+(y−b)²=r²: centre (a,b), radius r.

Exam questions — 4 questions · 10 marks · Edexcel 4MA1 style

3 marksStraight line equation4MA1 style ✏️ No calculator

Find the equation of the line passing through (2, 5) and (6, 13). Give your answer in the form y = mx + c. (3 marks)

Hint: Step 1: gradient = (y₂−y₁)/(x₂−x₁). Step 2: substitute gradient and one point into y−y₁ = m(x−x₁). Step 3: rearrange to y = mx+c.

+30 XP

3 marksPerpendicular linesGrade 7 Booklet style ✏️ No calculator

Line L has equation y = 3x − 2. Line M is perpendicular to L and passes through (6, 1). Find the equation of line M. (3 marks)

Hint: Perpendicular gradient = −1/m. Gradient of L is 3, so gradient of M is −1/3. Then use y−y₁ = m(x−x₁) with the given point.

+30 XP

3 marksGraph transformationGrade 7 Booklet style ✏️ No calculator

The graph of y = f(x) passes through (3, 4). (a) What are the coordinates of this point on y = f(x+2)? (b) On y = 2f(x)? (c) On y = f(−x)? (3 marks)

Hint: (a) f(x+2) shifts left by 2 — subtract 2 from x-coordinate. (b) 2f(x) doubles the y-coordinate. (c) f(−x) reflects in y-axis — negate x-coordinate.

+30 XP

1 markGradient of perpendicular4MA1 style

A line has gradient 2/5. What is the gradient of a line perpendicular to it?

A 5/2

B −2/5

C −5/2

D 2/5

1 markTransformation identification4MA1 style

The graph y = f(x−3) compared to y = f(x) is:

A Translated 3 units to the left

B Translated 3 units to the right

C Stretched by factor 3 in the x-direction

D Reflected in the line x = 3

Graphs & Coordinate Geometry

Straight line graphs

Graph transformations

Quadratic and other graphs

Module complete! 🎉