Video summary

Ultimate CCEA GCSE Maths M6 Revision Video - Corbettmaths

Main summary

Key takeaways

Educational

Main ideas / lessons conveyed

  • The video is an M6 (CCEA GCSE Maths) revision overview, aimed at helping students become familiar with all M6 topic areas.
  • It emphasizes that to do well in M6, students should already know M5 topics and also M1/M2 topics, especially those that commonly appear on the non-calculator paper (e.g., fraction operations).
  • Topic areas are grouped by colors:
    • Geometry (green): shape, space, measure
    • Algebra (blue): algebra topics
    • Statistics & probability (orange)
    • Number (red)
  • The presenter uses a “quick coverage” structure: ~3–4 minutes per topic, referring viewers to:
    • A checklist with links to more detailed Corbettmaths videos
    • An “M6 revision question booklet” (practice for every topic + QR-code answers)
    • Recommended revision card sets/booklets and a “short daily practice” method.

Methodologies / instructions

Bearings (3-figure bearings; clockwise from North)

Bearing of B from A

Given two towns A and B, you are asked for: bearing of B from A.

  1. Join the towns with a straight line.
  2. Draw a North line at the starting town (A).
  3. Use a protractor with 0° on the North line.
  4. Measure the clockwise angle from North to the line joining A → B.
  5. Record the 3-figure bearing (e.g., 105°, 044°).

Reflex bearing handling (>180° and <360°)

  • Option 1: Convert using the small angle
    • Measure the small interior angle (e.g., 35°).
    • Compute: bearing = 360° − small angle
      • Example: 360 − 35 = 325°
  • Option 2: Measure the reflex directly
    • Measure the reflex using a method such as placing 0° at the bottom and reading clockwise reflex value.
    • Example shown leading to 325° using straight-line reasoning (180° reference).
  • Option 3: Split the reflex using a straight-line reference
    • Draw a straight line/south reference and split the reflex into parts summing to the reflex bearing.

Back bearings (opposite direction)

  • If the given bearing is < 180°: back bearing = 180° + given bearing

  • If the given bearing is > 180°: back bearing = given bearing − 180°

  • Alternative method: co-interior angles on parallel north lines

    • Co-interior angles sum to 180°, then convert to reflex form by 360° − other angle.

Angles & polygons (interior/exterior angle rules)

Sum of interior angles of an n-sided polygon

  • Formula: (n − 2) × 180°
  • Example (12-sided): (12 − 2) × 180 = 1800°

Work backwards to find number of sides

  • If the interior sum is S: (n − 2) × 180 = S

  • Then: n = (S/180) + 2

Sum of exterior angles of a polygon

  • Sum of exterior angles (one per vertex) = 360°

Interior and exterior on a straight line

  • Interior + exterior = 180°

Regular polygons

  • Each exterior angle: 360° ÷ n
  • Each interior angle:
    • Either: find interior sum ((n − 2)×180) then divide by n
    • Or: use interior = 180 − exterior

Translations (vector notation)

  • Translation vector written as (horizontal, vertical), e.g. (−1, 5).

Meaning of vector components

  • Top number = horizontal move
    • Positive → move right
    • Negative → move left (by the absolute value)
  • Bottom number = vertical move
    • Positive → move up
    • Negative → move down

Procedure

  1. For each vertex, move it by the vector:
    • Left/right by the horizontal component
    • Up/down by the vertical component
  2. Join the translated points to form the new shape.

Rotations (about a point not necessarily the origin)

  • Rotation is specified by:
    • Angle (e.g., 90°)
    • Direction (clockwise/anticlockwise)
    • Center of rotation (e.g., (2, −1))

Procedure using tracing paper

  1. Mark the center of rotation.
  2. Place tracing paper so its center point aligns with the marked center.
  3. Rotate the tracing paper by the given angle (anticlockwise in the example).
  4. Trace/mark where the original shape lands.
  5. Redraw for clarity and join corresponding vertices.

Reflections (in lines x = a or y = b)

Mirror lines

  • Mirror line x = a
    • Vertical line through x = a.
    • Every point on the line has x-coordinate a.
  • Mirror line y = b
    • Horizontal line through y = b.
    • Every point on the line has y-coordinate b.

Reflection rule

  • For a point, measure perpendicular distance to the mirror line, then move the same distance to the other side.

Procedure

  1. Identify the mirror line (x = constant or y = constant).
  2. For each vertex, measure its distance from the mirror line and place the reflected vertex symmetrically.
  3. Join reflected points to form the reflected shape.

Enlargements (scale factor from a center)

  • Defined by:
    • Scale factor k (e.g., 3)
    • Center of enlargement (e.g., (3, 2))

Core rule

  • Each vertex moves k times further away from the center.

Procedure

  1. Plot the center.
  2. For each vertex:
    • Determine its horizontal/vertical offset from the center.
    • Multiply offsets by k.
    • Place the new vertex at the scaled offset location.
  3. Connect new vertices.

Perimeter and area scaling

  • With scale factor k:
    • Perimeter becomes k times larger
    • Area becomes k² times larger
  • Example correction:
    • Scale factor 3 → area becomes 9 times larger (not 3 times).

Constructions (compass & straightedge)

1) Perpendicular bisector of segment AB

  • A line that is perpendicular (90°) to AB and cuts AB in half.

Steps

  1. Set compasses to a radius reaching beyond both endpoints.
  2. With compass centered at A, draw an arc above/below AB.
  3. With compass centered at B (same radius), draw a matching arc.
  4. Draw a straight line through the arc intersection points.
  5. Result: perpendicular bisector.

2) Angle bisector of angle ABC

Steps

  1. Draw arcs from A and C with the same compass radius, intersecting near the angle.
  2. Draw a line from B to the intersection point(s).
  3. Result: bisects angle ABC.

3) Line perpendicular to AB passing through point C

  • Make two arcs centered at C that intersect AB at two points, then construct the perpendicular bisector of that segment.
  • The perpendicular bisector passes through C and is perpendicular to AB.

4) Perpendicular to AB at point C (two-stage approach)

  • The video describes a compass-based auxiliary segment approach, then taking its perpendicular bisector (same outcome).

5) Construct an equilateral triangle on a given segment

Steps

  1. Set compass radius to the given side length.
  2. Draw an arc from endpoint A with that radius.
  3. Draw an arc from endpoint B with the same radius.
  4. Arc intersections give the third vertex.
  5. Join all three points. - Result: equilateral triangle (all sides equal; angles 60°).

Loci (points at a fixed distance from a line/point)

1) Locus of points 2 cm from a line

  • Points are all positions at distance 2 cm, measured perpendicularly to the line.

Key idea (as described)

  • Sketch points at distance 2 cm above the line and below the line.
  • At line endpoints, the locus becomes semicircles with radius 2 cm around each endpoint.

2) Locus intersection for multiple distance constraints

  • Example approach:
    • Draw a circle/region for points 8 miles from A
    • Draw a second for points 5 miles from B
    • The solution is where the conditions intersect.

Congruence

  • Congruent shapes have:
    • Exactly the same size and shape
    • Corresponding sides and angles equal (as described in the video)
  • They appear with matching geometry features (e.g., equal base/height for right triangles).

Laws of indices (algebraic power rules)

  • Same base multiplication: ( a^m \times a^n = a^{m+n} )
  • Same base division: ( a^m \div a^n = a^{m-n} )
  • Power of a power: ( (a^m)^n = a^{m\times n} )

Examples

  • ( y^8 \times y^3 = y^{11} )
  • ( y^{15} \div y^5 = y^{10} )
  • ( (y^6)^2 = y^{12} )

Trial and improvement (solve to 1 decimal place)

  • Target example: solve an equation like (x^3 + 7x = 30) to 1 d.p.

Procedure used

  1. Choose a starting value (e.g., 2) and substitute into the LHS.
  2. Compare to 30:
    • Too low → increase (x)
    • Too high → decrease (x)
  3. Bracket the solution:
    • Find one value giving low and one giving high (e.g., 2.3 low, 2.4 high).
  4. Use a midpoint check (e.g., 2.35) to decide which side it falls on.
  5. Decide which candidate is closer to the true solution. - Final answer in the example: (x = 2.4) (to 1 d.p.).

Solving inequalities (algebra steps + inequality direction)

General method

  • Apply the same algebra operations to both sides.
  • If you multiply or divide by a negative number, the inequality flips.

Examples shown

  • 5x > 30
    • Divide by 5 → x > 6
  • 3x + 4 ≤ 31
    • Subtract 4 → 3x ≤ 27
    • Divide by 3 → x ≤ 9
  • 8x + 1 < 10x − 6
    • Subtract 8x → 1 < 2x − 6
    • Add 6 → 7 < 2x
    • Divide by 2 → 3.5 < x
    • Rewrite carefully as x > 3.5

Inequalities with number lines

  • Hollow circle: strict inequality (< or >)
  • Filled circle: inclusive inequality (≤ or ≥)
  • Arrow:
    • Right for “greater than”
    • Left for “less than”
  • Combined inequalities (e.g., 1 < x ≤ 3):
    • Hollow at 1, filled at 3, shade/line between.

Changing the subject (make a different variable the subject)

  • Example: start with ( t = aw - c ) and make w the subject.

Steps

  1. Add (c) to both sides:
    • ( t + c = aw )
  2. Divide by (a):
    • ( w = \frac{t + c}{a} )

nth term of a sequence

Method

  1. Determine whether the sequence goes up or down by a constant amount.
  2. Write the basic multiple sequence (e.g., multiples of 5 or 2).
  3. Adjust with a constant offset to match the given sequence.

Examples

  • 3, 8, 13, 18, … increases by 5
    • Base: (5n) → gives 5, 10, 15, 20
    • Need correction −2
    • nth term: (5n - 2)
  • 7, 9, 11, 13, … increases by 2
    • nth term: (2n + 5)
  • 15, 12, 9, 6, … decreases by 3
    • nth term: (-3n + 18)

Using nth term

  • To find (e.g.) the 100th term: substitute (n=100).
  • To check if a number is in the sequence:
    • Set nth term equal to that number and solve for (n).
    • If (n) is a whole number, it appears.

Simultaneous equations (graphical approach)

  • Draw both graphs (e.g., (y = 3 - x) and (y = 2x - 3)).
  • Find where they intersect.
  • Intersection coordinate (x, y) gives the solution.

Quadratic graphs and solving graphically

Shape of quadratics

  • Parabolas:
    • Positive (x^2) coefficient → “U” shape
    • Negative (x^2) coefficient → “∩” shape

Completing an xy table and plotting

  • For ( y = x^2 + x - 4 ):
    • Substitute x values to get y values,
    • Plot points,
    • Draw a smooth curve.

Solving quadratics graphically

  • To find x when (y =) a constant:
    1. Draw the horizontal line (y =) that constant.
    2. Find intersection points with the parabola.
    3. Estimate x-values from the graph (not exact).

Probability & statistics (outcomes, sample spaces, experimental probability, sampling)

1) Listing outcomes (two dice; add scores)

  • Totals range from 2 to 12.
  • Outcomes correspond to all achievable sums from two dice.

2) Sample space table for multiplying choices from two bags

  • Create a table of combinations:
    • Bag 1 options × Bag 2 options
  • Identify outcomes satisfying a condition (e.g., “multiple of 4”).
  • Probability:
    • P = favourable outcomes / total outcomes

3) Relative frequency / experimental probability

  • Relative frequency:
    • (number of times it happened) / (total trials)
  • Example given:
    • (P(a)=\frac{4}{7}), (P(b)=\frac{2}{7}), (P(c)=\frac{1}{7})
  • Spinner example:
    • expected successes = trials × relative frequency

4) Sampling (representativeness)

  • Ensure the sample is large enough and representative.
  • Avoid bias (e.g., selecting only top achievers).

Binary numbers (conversion between binary and decimal)

Binary → decimal

  1. Write column headings as powers of 2: 1, 2, 4, 8, 16, …
  2. Multiply each 1 digit by its power of 2 and add. - Examples:
    • (1101_2 = 8 + 4 + 1 = 13)
    • (10110_2 = 16 + 4 + 2 = 22)

Decimal → binary

  1. Choose the largest power of 2 ≤ the decimal number.
  2. Subtract it and continue with descending powers.
  3. Put 1 where a power is used; 0 where it isn’t. - Examples:
    • 18 → 16 + 2 → 10010
    • 27 → 16 + 8 + 2 + 1 → 11011

Sources / speakers featured

  • Speaker/Presenter: The channel host speaking throughout (identified in subtitles as “Corbettmaths” / “Corbettmavs”).
  • References/Resources (not additional speakers):
    • Corbettmaths (videos and checklist links)
    • CCEA GCSE Maths M6 Revision Question Booklet
    • “Corbettmaths revision cards” (foundation sets/booklets mentioned)

Original video