Formula List NUMBERS 1. Natural number (π΅): counting numbers. Example: 1 , 2 , 3 , 4 , … … 2. Integer (β€): Positive and negative whole numbers. … … − 3 , −2 , −1 , 0 , 1 , 2 , 3 … … Example: β€− β€+ Neutral Integer 3. Real number (β): 1 −3.5 , −2 , 0 , 2 Can be placed in number line. Example: , π , √2 … … 4. Unreal number: Cannot be placed in the number line 5. Square number: 1, 4, 9, 25, 36, 49, … … 6. Cubic number: 1, 8, 27, 64, 125, 216, 343 … … 1 0 Example: √−2 , , ππ‘π 7. Prime number (π·): β Integer starts from 2 β Divisible by 1 & itself β Has exactly two factors. Note: 1 is not a prime number. Example: 2 , 3 , 5 , 7 , 11 , 13 , 17, 19 … … Factor of 6 8. Factors of π: 1, 2, 3, 6 9. Multiples of π: 6, 12 , 24 , 36 , 48 , 60, … … 10. Even numbers: −4, −2, 0 , 2 , 4 , 6 , 8 , … … 11. Odd numbers: −5, −3, −1, 1 , 3 , 5 , 7 , 9, … …. 1 , 2 , 3 , 6 , 12 , 18 … … Multiple of 6 π 12. Rational Number (πΈ): A number that can be converted into fraction of integers (π ) . Example: 0 0 =1 2.5 = 25 10 −5.1 =− 51 10 √16 4 =1 π 13. Irrational Number (πΈ′ ): A number that cannot be converted into fraction of integers (≠ π ) . 14. Product of prime factors: [Example] 2 48 2 24 2 12 2 6 3 ∴ 48 = 2 × 2 × 2 × 2 × 3 = 24 × 3 2 80 2 40 2 20 2 10 5 ∴ 48 = 2 × 2 × 2 × 2 × 5 = 24 × 5 Formula List SEQUENCE β Type – 1 (Consecutive numbers): π»π +2 π»π π»π 4 5 π»π π»π −1 0 3 π»π −3 −2 π»π ….. +2 π+2 π»π ….. −3 π−3 β Type – 2 (Common Difference): π»π π»π π»π π»π 7 9 11 13 2 2 π»π π + (π − 1)π 2 = 7 + (π − 1) × 2 π»π π»π π»π π»π π»π 6 10 16 24 34 4 6 2 8 2 10 2 π»π π + (π − 1)π1 + (π−1)(π−2) = 6 + (π − 1) × 4 + 2 π2 (π−1)(π−2) 2 ×2 Formula List GEOMETRY Acute Angle 0 < π < 90° Obtuse Angle 90° < π < 180° Reflex Angle 180° < π < 360° Right Angle π = 90° π π π π External ∠ of β Triangle Allied Angle Parallelogram Quadrilateral π π π=π ∠π centred at a point Vertically opposite angles Adjacent ∠π on a straight line Adjacent ∠π on a right angle Corresponding angles Alternate angles Formula List CIRCLE THEOREM π΄ π¦° π₯° π₯° 2π¦° π§° .π .π 2π₯° π΄ π΅ πΆ π΅ π¦° π΄ π΅ ∠π΄πΆπ΅ = 90° ∠π΄ππ΅ = 2π₯ π₯° = π¦° = π§° Reflex ∠π΄ππ΅ = 2π¦ π΅ π΄ π΅ π΄ π° π° π° π° π° π· π· .π π° πΆ π₯° π΄ πΆ π° + π° = 180° π΅ πΆ ∠π΄πΆπ = ∠π΅πΆπ = 90° π° = π₯° π° + π° = 180° π΄ π΄ .π π΅ π° π° π΅ π π₯° π¦° π π° = π¦° π° = π₯° πΆ π π΄π΅ = π΅πΆ Formula List LOCUS CONSTRUCTION 1. Locus of a point π·, equidistant from π¨ and π©: π· π© π¨ ← Locus ∴ Perpendicular bisector is the locus of a point π, equidistant from π΄ and π΅. 2. Locus of a point π·, equidistant from π¨π© and π©πͺ: π¨ ← Locus π· π© πͺ ∴ Angle bisector is the locus of a point π, equidistant from π΄π΅ and π΅πΆ. 3. Locus of a point π·, so that it is π cm from π·: π·′ π cm ← Locus π· ∴ Circle is the locus of a point π, 4 cm from it. Formula List 4. Locus of a point π·, so that it is always π cm from a line π¨π©: π· π cm π cm π cm π© π¨ π cm Locus ∴ Capsule is the locus of a point π, so that π is always 3 cm from π΄π΅. Μ π© is always ππ°: 5. Locus of a point π·, so that π¨π· π· π¨ Locus π© ∴ Circle is the locus of a point π, so that π΄πΜ π΅ is always 90°. 6. Locus of a point π·, so that triangle π¨π·π© always has an area of ππ ππ¦π : Locus π· π· π cm π cm π¨ π© π cm π· 1 Locus ∴ Parallel lines π cm from π΄π΅ are the locus of a point π, so that triangle π΄ππ΅ is always 2 × 8 × 3 = 12 cm2. Formula List SIMILARITY 1. If three angles equal, triangles are similar. 2. If two angles equal, 3rd angle automatically is equal. 3. Opposite sides of equal angles are corresponding sides, π1 and π2 . 4. π1 π2 5. π΄1 π΄2 6. π = π1 2 π 2 π2 π1 π1 3 π1 π = ( ) , π = π1 π2 π2 2 2 π΄ 3 π 2 = ( 1) 1 1 2 2 π€ = π€1 2 7. (π΄ ) = (π ) CONGRUENCY POLYGON A regular polygon has all sides equal and all interior angles equal For a π-sided polygon: Sum of all exterior angles (Regular + Irregular polygon) 360° Sum of all interior angles (Regular + Irregular polygon) (π − 2) × 180° Each exterior angle (Regular polygon only) Each interior angle (Regular polygon only) Sum of an exterior angle and an interior angle 360° π (π − 2) × 180° π (π − 2) × 180° 360° + = 180° π π Formula List Naming of polygons: 3-sided: Triangle 4-sided: Quadrilateral 5-sided: Pentagon 6-sided: Hexagon 7-sided: Heptagon 8-sided: Octagon 9-sided: Nonagon 10-sided: Decagon TRIGONOMETRY SOH CAH TOA sin π = πππππ ππ‘π π πππ π = π»π¦πππ‘πππ’π π π» π΄ cos π = π΄πππππππ‘ π πππ π΄ = π»π¦πππ‘πππ’π π π» Opposite For a right-angled triangle, π¨π©πͺ: π πππππ ππ‘π π πππ π tan π = = π΄πππππππ‘ π πππ π΄ π΅ πΆ Adjacent SINE RULE π΄ π π π = = sin π΄ sin π΅ sin πΆ π π΅ π π πΆ COSINE RULE To find the length of a side π2 = π 2 + π 2 − 2ππ cos π΄ π 2 = π2 + π 2 − 2ππ cos π΅ To find an angle when all three sides are given cos π΄ = π 2 + π 2 − π2 2ππ cos π΅ = π2 + π 2 − π 2 2ππ cos πΆ = π2 + π 2 − π 2 2ππ π 2 = π2 + π 2 − 2ππ cos πΆ Formula List BEARING The bearing of a point π΅ from another point π΄ is: • an angle measured from the north at π΄, • in a clockwise direction. π π΅ π π΅ 050° 230° Example: The bearing of π΅ from π΄ is 050°. The bearing of π΄ from π΅ is 230°. π΄ π΄ ELEVATION – DEPRESSION 1. Angle of Elevation → line of sight above horizontal line. 2. Angle of Depression → line of sight below horizontal line. Angle of elevation of π» from π΄ = π½° Angle of depression of π΄ from π» = π½° Formula List ARITHMETIC β Ratio: If the amount π΄ is divided between π, π and π in the ratio π: π: π Share of π = π π+π+π × π΄, Share of π = π π+π+π ×π΄ Share of π = π π+π+π ×π΄ β Percentage: loss 1. Profit = sale price – cost price 4. Percentage loss = cost price × 100 2. Loss = cost price – sale price increase 5. Percentage increase = original value × 100 profit 3. Percentage profit = cost price × 100 π΄ 6. % of π΄ compared with π΅ = π΅ × 100 β Simple Interest: Here, ππ π πΌ= 100 πΌ = Interest π = Principle β Compound Interest: π = Time (Year) π΄ = π (1 + π ) 100 π π = Rate of Interest π΄ = Final Amount β Speed, Distance, Time: 1. • π =π ×π‘ • π = • π‘= π π‘ π π 2. π π π‘ ππ¬π part • π = π₯ km/h • π‘ = π‘1 h • π = π₯π‘1 ππ§π part • π = π¦ km/h • π‘ = π‘2 h • π = π¦π‘2 Avg. speed = = Total distance Total time π₯π‘1 +π¦π‘2 π‘1 +π‘2 β Map Scale: 1. Scale of a map = (map distance): (actual distance) 2. Scale of a map is written in the form 1: π, where 1 is map distance ratio and π is actual distance ratio. π 1 3. π1 = π where π1 = map distance and π2 = actual distance. 2 π΄ 1 2 4. π΄1 = (π) 2 π 1 3 5. π1 = (π) 2 where π΄1 = map area and π΄2 = actual area. where π1 = model volume and π2 = actual volume. Formula List SETS 1. Notation: Number of elements in set π΄ π(π΄) π΄ = {π, π, π, π } ∴ π(π΄) = 4 Intersection ∩ π΄ = {π, π, β, π}, π΅ = {π, β, π, π} ∴ π΄ ∩ π΅ = {π, β} Union ∪ π΄ = {π, π, β, π}, π΅ = {π, β, π, π} ∴ π΄ ∪ π΅ = {π, π, β, π, π, π} Universal set π Complement of set π΄ π΄′ π΄ ∪ π΄′ = π π = {1, 2, 3, 4}, π΄ = {2, 3} ∴ π΄′ = {1, 4} Null set/Empty set π π΄ = {π₯, π¦, π§}, π΅ = {2, 3, 4} ∴ π΄∩π΅ = π ∴ π(π΄ ∩ π΅) = 0 Is an element of ∈ π΄ = {π₯, π¦, π§} π₯ ∈ π΄, 2 ∉ π΄ Is a subset of ⊂ π΄ = {π₯, π¦, π§}, π΅ = {π¦, π₯}, πΆ = {π₯, π} ∴ π΅ ⊂ π΄ and πΆ ⊄ π΄ π 2. All possible subsets: • • If π¨ = {π, π, π}, subsets of π¨ are: {π}, {π}, {π}, {π, π}, {π, π}, {π, π}, {π, π, π}, π. Both π and π¨ are subsets of π¨. 3. De Morgan’s rules: π΄′ ∩ π΅′ = (π΄ ∪ π΅)′ π΄′ ∩ π΅′ ∩ πΆ ′ = (π΄ ∪ π΅ ∪ πΆ)′ π΄′ ∪ π΅′ = (π΄ ∩ π΅)′ π΄′ ∪ π΅′ ∪ πΆ ′ = (π΄ ∩ π΅ ∩ πΆ)′ 4. Subset: Isosceles Rightangled Acute Equilateral Scalene Obtuse Formula List π = ππππ¦ππππ • • • • • • Rectangle Rhombus Parallelogram Square Kite Quadrilateral Trapezium Rhombus → all sides equal. Rectangle → all angles equal. Square → all angles 90°, all sides equal. Parallelogram → opposite sides parallel. Kite → one line symmetry. Trapezium → one pair of parallel sides. Triangle π = ππ’πππππ Real, β Rational, β Integer, β€ • Natural, β • 1, 2, 3, … • • … , −1, 0, 1, … β is the set of natural numbers or positive integers {1, 2, 3, 4, … … }. β€ is the set of integers {… , −2, −1, 0, 1, 2, … }. β is the set of rational numbers. β is the set of real numbers. ½, ¾, 2.5 π, ln 2, √3 5. Description of a Venn diagram: • • • • • • • • • • • • • • 1 → ππππ¦ π΄. 2 → ππππ¦ π΅. 3 → ππππ¦ πΆ. 4 → ππππ¦ π΄ πππ π΅ ππ ππ’π π‘ π΄ πππ π΅ ππ π΄ πππ π΅ ππ’π‘ πππ‘ πΆ. 5 → ππππ¦ π΄ πππ πΆ ππ ππ’π π‘ π΄ πππ πΆ ππ π΄ πππ πΆ ππ’π‘ πππ‘ π΅. 6 → ππππ¦ π΅ πππ πΆ ππ ππ’π π‘ π΅ πππ πΆ ππ π΅ πππ πΆ ππ’π‘ πππ‘ π΄. 7 → π΄ππ π΄, π΅ πππ πΆ. 8 → πππ‘ π΄, π΅ πππ πΆ. 4, 7 → π΅ππ‘β π΄ πππ π΅ ππ ππ‘ ππππ π‘ π΄ πππ π΅. 5, 7 → π΅ππ‘β π΄ πππ πΆ ππ ππ‘ ππππ π‘ π΄ πππ πΆ. 6, 7 → π΅ππ‘β π΅ πππ πΆ ππ ππ‘ ππππ π‘ π΅ πππ πΆ. 1, 4, 5, 7 → π΄. 2, 4, 6, 7 → π΅. 3, 5, 6, 7 → πΆ. π π΄ 4 1 5 8 7 3 πΆ 2 6 π΅ Formula List MATRIX 1. Multiplication by a scalar π ( π π ( π π ( π 2. Addition of matrix 3. Subtraction of matrix 4. Multiplication of matrix π π ππ ππ π( )=( ) π π ππ ππ π€ π₯ π+π€ π+π₯ π ) + ( π¦ π§) = ( ) π+π¦ π+π§ π π€ π₯ π−π€ π−π₯ π ) − ( π¦ π§) = ( ) π−π¦ π−π§ π ππ€ + ππ¦ ππ₯ + ππ§ π π€ π₯ )( )=( ) ππ€ + ππ¦ ππ₯ + ππ§ π π¦ π§ 5. If two matrices have order (π × π) and (π × π ), they can be multiplied if π = π and the order of the resultant matrix will be (π × π ). π€ π₯ π π ( ) = ( π¦ π§) π π ∴ π = π€, π = π₯, π = π¦, π = π§. 6. Equal matrix π π 7. Determinant of ( ) is ππ − ππ. π π 1 π π π −π 8. Inverse of ( ) is ( ). If ππ − ππ = 0, the matrix has no inverse. ππ−ππ π π −π π 9. Identity matrix: (1 × 1) (2 × 2) (3 × 3) 1 0 0 π = (0 1 0) 0 0 1 10. If any matrix is multiplied by identity matrix, the matrix remains same, i.e., π × π = π. The product of inverse matrix and original matrix is identity matrix i.e., π−1 × π = π, 1 0 where π = ( ). 0 1 π = (1) 1 0 π=( ) 0 1 MATRIX TRANSFORMATION Formula List Formula List 5. Identification of transformation and their fully description: Transformation Identification Need to describe fully Enlargement Change of area or change of side-length Centre and scale factor Translation Change of position without rotation Column vector Reflection 180° rotation Corresponding object – image joining lines are parallel Corresponding object – image joining lines intersects at the same point, which is the centre of 180° rotation Other rotation - 6. πππ° rotation is same as enlargement, πΊ. π. = −π. 7. How to find unknown matrix? Equation of reflection line Centre Angle, direction and centre Formula List Case – 2: For known transformation, use the following matrices. Transformation Rotation 90°, anticlockwise, centre (0, 0) Rotation 90°, clockwise, centre (0, 0) Rotation 180°,centre (0, 0) Reflection in π₯ −axis Reflection in π¦ −axis Reflection in π¦ = π₯ Reflection in π¦ = −π₯ Enlargement, scale factor π, centre (0, 0) Matrix 0 −1 ( ) 1 0 0 1 ( ) −1 0 −1 0 ( ) 0 −1 1 0 ( ) 0 −1 −1 0 ( ) 0 1 0 1 ( ) 1 0 0 −1 ( ) −1 0 π ( 0 0 ) π 8. πππ‘πππ₯ × ππππππ‘ = πΌππππ 9. πππ(β, π) means (β, π) is transformed by π, followed by π, followed by π. Case – 3: Formula List VECTOR 1 2 Column Vector: π₯1 • πβ = (π¦ ) or π₯1 π’ + π¦1 π£ 1 6 βββββ π΄π΅ −−−/−−− Step 1: ππ = −−−/−−− = π (constant) βββββ • |πβ| = √π₯1 2 + π¦1 2 • Unit Vector of πβ: |πβ| • ∠ between πβ and π₯-axis: tan π = • π₯1 π₯2 π₯ +π₯ πβ + π = (π¦ ) + (π¦ ) = (π¦1 + π¦2 ). • π₯1 ππ₯ ππβ = π (π¦ ) = ( 1 ). ππ¦1 1 Step 2: ∴ π΄π΅ β₯ ππ β π 1 Show that π¨π© β₯ π·π»: Use the following steps: 2 1 π¦1 π₯1 7 2 Show π¨, π© and πͺ lie on a straight line (collinear). βββββ π΄π΅ −−−/−−− Step 1: βββββ = −−−/−−− = π (constant) π΅πΆ Step 2: ∴ π΄π΅ β₯ π΅πΆ Step 3: Since π΅ is common, therefore π΄, π΅ and πΆ lie on a straight line. 3 • • ππ ππ If (π ) = (π ), π 8 π o π₯1 = π₯2 ------------ (i) o π¦1 = π¦2 ------------ (ii) ππ ππ If (π ) || (π ), π π π₯1 π₯2 o (π¦ ) = π (π¦ ) 1 2 o π₯1 = ππ₯2 ------------ (i) o π¦1 = ππ¦2 ------------ (ii) o π₯ • 5 • • • 2 9 β π΄(β, π) means βββββ ππ΄ = ( ). π In terms of position vector: βββββ = ππ΅ βββββ − ππ΄ βββββ o π΄π΅ o ∴ π΄π΅πΆπ· is a parallelogram. Or prove βββββ π΄π· = βββββ π΅πΆ ∴ π΄π΅πΆπ· is a parallelogram. ⇒ π₯πβ + π¦π = π(ππβ + ππ) ∴ π₯ = ππ ------------ (i) ∴ π¦ = ππ ------------ (ii) ββββββ For two geometrical properties of π¨π© and ββββββ π·πΈ: Use the following steps: Step 1: Ratio • βββββ ππ = ββββββ ππ − βββββ ππ βββββ = (5). Example: Given π΄(3, 2) and π΄π΅ 2 βββββ Find ππ΅. 5 • ⇒ βββββ ππ΅ − βββββ ππ΄ = ( ). 2 3 8 5 βββββ • ∴ ππ΅ = ( ) + ( ) = ( ). 2 4 2 Show that π¨π©πͺπ« is a parallelogram: βββββ = π·πΆ βββββ Either prove π΄π΅ π¦ π βββββ = π × π΄πΆ βββββ Or otherwise, π΄π΅ π₯ Position Vector: • π₯ π Step 2: Use shortcut = to find π. Shortcut: π¦1 = π¦2 1 4 Given π¨, π© and πͺ are collinear (or parallel), find π. Step 1: Find βββββ π΄π΅ = π₯πβ + π¦π and βββββ π΄πΆ = ππβ + ππ. βββββ π΄π΅ βββββ ππ = π π (constant) Step 2: Properties • βββββ π΄π΅ β₯ βββββ ππ . • 10 (ii) π΄π΅: ππ = π: π. Ratio example: βββββ = π© β + 2πͺ β. Given π΄π΅: π΅πΆ = 5: 3 and π΄π΅ Find βββββ π΄πΆ . βββββ π΄π΅ 5 • ⇒ βββββ = 5+3 • ⇒ • βββββ = π© + ∴ π΄πΆ 5 π΄πΆ π©+2πͺ π΄πΆ = 8 5 8 16 πͺ 5 Formula List 11 Area Ratio: 1 β π¨π©π« 2 × 2 × β = β π©π«πͺ 1 × 3 × β 2 1 β π¨π·πΈ 2 × 3 × 1 × sin π = β π¨π©πͺ 1 × 4 × 2 × sin π 2 β π·πΊπ» 3 2 =( ) β π·πΈπΉ 5 PROBABILITY π = number of π΄, π‘ = number of π΅, π’ = number of πΆ, π = total number. π π‘ π’ 1. π(π΄) = π, π(π΅) = π, π(πΆ) = π π π π π 2. (i) π(π΄π΄) = × , [If replaced] π π π 3. (i) π(π΄π΄π΄) = π × π × π [If replaced] π π‘ 4. (i) π(π΄π΅) = π × π × 2 [If replaced] π π (ii) π(π΄π΄) = × π π −1 π−1 [If not replaced] π −1 π −2 (ii) π(π΄π΄π΄) = π × π−1 × π−2 [If not replaced] π π‘ (ii) π(π΄π΅) = π × π−1 × 2 [If not replaced] 5. π΄, π΅ and πΆ are arranged in the following 6 ways: π΄π΅πΆ, π΄πΆπ΅, π΅π΄πΆ, π΅πΆπ΄, πΆπ΄π΅, πΆπ΅π΄. π π‘ π’ 6. (i) π(π΄π΅πΆ) = π × π × π × 6 [If replaced] π π‘ π’ (ii) π(π΄π΅πΆ) = π × π−1 × π−2 × 6 [If not replaced] 7. For 2 draws, π(diferent color) = 1 − π(same color). 8. AND rule: π(π΄ and π΅) = π(π΄) × π(π΅). 9. OR rule: π(π΄ or π΅) = π(π΄) + π(π΅). 10. π(π΄) + π(π΅) + π(πΆ) = 1. 11. The probabilities of disc, ball, bead, sweet etc. may change or not according to replacement. But probabilities of die, coin etc. always remain unchanged. Formula List MENSURATION 1. Triangles π΅ π΅ π΄ πΆ π π΄ πΆ π π π 1 1 Area of βπ΄π΅πΆ = × ππ × sin π Area of βπ΄π΅πΆ = × π × β 2 2 2. π π β β SINE RULE π΄ π sin π΄ π π π΅ πΆ π = π sin π΅ = π sin πΆ COSINE RULE For sides: π2 = π 2 + π 2 − 2ππ cos π΄ For angles: cos π΅ = π 2 +π 2 −π2 2ππ 3. Area of Quadrilateral π β2 π β1 π β π Parallelogram Rhombus i) π΅ππ π × βπππβπ‘ π. π (π × β1 ) ππ (π × β2 ) i) πππ π × βπππβπ‘ ii) ππ sin π ii) π2 sin π Kite Trapezium 1 1 π΄πππ = (π + π) × β π΄πππ = × π1 × π2 2 2 1 iii) π1 π2 2 4. Circle Length Area r π = 2ππ Major arc Minor arc r π= 2ππ 2 π= 2ππ 4 π΄ = ππ 2 π΄= ππ 2 2 π΄= ππ 2 4 Formula List 5. Sector and Arc πππππ πππ = π₯ 360 πππππ π πππ‘ππ = × 2ππ π₯ 360 × ππ 2 πππππ πππ = π₯ π π 360 πππππ π πππ‘ππ = × 2ππ π 360 × ππ 2 6. Area of polygon: Step 1: π = 360 π 1 Step 2: ππππ¦πππ ππππ = [2 × π × π × sin π] × π π». πΊ. π¨ = π»ππππ πππππππ ππππ 7. Cylinder (solid) πͺ. πΊ. π¨ = πͺπππππ πππππππ ππππ Cone (solid) Sphere Hemisphere (solid) π π β β π π π π ππππ’ππ = ππ 2 β π = √β2 + π 2 πΆ. π. π΄ = 2ππβ π. π. π΄ = 2ππβ + 2ππ 2 4 π£πππ’ππ = ππ 3 ππππ’ππ = ππ 2 β πΆ. π. π΄ = 4ππ 2 πΆ. π. π΄ = 2ππ 2 πΆ. π. π΄ = πππ π. π. π΄ = πππ + ππ 2 π. π. π΄ = 4ππ 2 π. π. π΄ = 2ππ 2 + ππ 2 3 1 Pyramid 2 ππππ’ππ = ππ 3 3 3 Cuboid Cube π π· π΅ πΆ π π΄ π΄ π΅ (Tetrahedron) ππππ’ππ = 1 3 × π΅ππ π ππππ × β π. π΄ = π΅ππ π + πππππ‘ ππ’πππππ ππππ’ππ = π × π × β π. π. π΄ (πΆπππ ππ) = 2ππ + 2πβ + 2πβ ππππ’ππ = π 3 π. π. π΄(ππππ ππ) = 6π 2 Formula List Prism π π Cross Sections: π½πππππ ππ πππππ = (ππππ ππ πππππ πππππππ) × ππππππ 8. Volume of Liquid issued (speed in m/s) ππ 1 π ππ = πΆ. π × π ππππ ππ 1 ππππ’π‘π = πΆ. π × π ππππ × 60 COORDINATE GEOMETRY (i) Length π΄π΅ = √(π₯2 − π₯1 )2 + (π¦2 − π¦1 )2 π π₯1 +π₯2 π¦1 +π¦2 , ) 2 2 (ii) Midpoint = ( (iii) Gradient = (iv) Co-ordinates of π = ( (v) Equation of π΄π΅, • π π¦2 −π¦1 π₯2 −π₯1 ππ₯2 +ππ₯1 ππ¦2 +ππ¦1 , π+π ) π+π π΅ (π₯2 , π¦2 ) π π΄ (π₯1 , π¦1 ) π¦ − π¦1 = π(π₯ − π₯1 ) β To find gradient: Case-1: Case-2: Case-3: π΅ (π₯2 , π¦2 ) 2 • π΄ 3 π΄ (π₯1 , π¦1 ) π¦ −π¦ Gradient, π = π₯2 −π₯1 2 m= 1 (vi) For parallel lines, Rise Run 2 ππ¦ =3 mat π΄ = ππ₯ (vii) For perpendicular lines, π1 = π2 π1 × π2 = −1 π1 π2 π2 1 ∴ π2 = − π 1 π1 Formula List β To find equation: Case-1: Case-2: π¦ π¦ (3, 5) • • 5 π = ππ + π π1 π¦=5 For π1 βΆ π = 1, π = 1 π₯=3 π₯ = −2 1 2 1 ∴ π¦ = 2π₯ + 1 π₯ 0 For π2 βΆ π = 1, π = −1 −2 3 0 π₯ ∴ π¦ = −π₯ + 1 π2 βͺ How to know if a point lies on the line or not Case-3: π¦ • (4, 8) does not lie on the line 3π₯ + 7π¦ = 18 π • (π₯1 , π¦1 ) 6 as when π₯ = 4, π¦ = 7 (≠ 8) • (6, 0) lies on the line as when π₯ = 6, π¦ = 0 π₯ π¦ − π¦1 = π(π₯ − π₯1 ) (viii) πΆ π΅ (π₯1 , π¦1 ) Area = • π΄ (π₯2 , π¦2 ) = • 1 |π₯ 2 3 π¦3 1 |( β ) 2 π΅ π₯2 π¦2 π΄ π₯1 π¦1 πΆ π₯3 | π¦3 − ( β )| • πΆ (π₯1 , π¦1 ) π¦ (x) π ππ₯ + ππ¦ = π 0 π π₯ For π· For πΈ π¦=0 ππ₯ + (π × 0) = π π π₯=π π₯=0 (π × 0) + ππ₯ = π π π¦=π Formula List (xi) − − − −(i) For intersecting point π, • Solve equations (i) and (ii) π − − − −(ii) (xii) Tangent, ππ¦ π π = ππ₯ (π₯1 , π¦1 ) Normal, 1 ππ = − π π Equation of tangent: π¦ − π¦1 = π π (π₯ − π₯1 ) Equation of normal: π¦ − π¦1 = ππ (π₯ − π₯1 ) FUNCTION 1. π π π π π π π π π π π π π π One to Many Many to One One to One π π 2. Vertical line test: π¦ π π¦ π¦ π₯ π₯ π π₯ π One to one Many to one One to many Function Function Not Function Inverse Inverse does not exist π/π΄ Formula List 3. π(π) to π−π (π) 4. π(π₯) = π Step – 1: Let π(π₯) = π¦ Step – 2: π¦ → π₯, π₯ → π¦ Step – 3: Make π¦ subject Step – 4: π¦ = π −1 (π₯) π₯ = π −1 (π₯) 5. π· of π(π₯) = π of π −1 (π₯) π of π(π₯) = π· of π −1 (π₯) 6. 7. π(π₯) = 2π₯ + 2 ππβ(π₯) π(π₯) = 4π₯ = ππ(π₯ 3 ) β(π₯) = π₯ 3 = π(4π₯ 3 ) = 2(4π₯ 3 ) + 2 STATISTICS β Histogram: Q1 50 < π€ ≤ 60 60 < π€ ≤ 75 75 < π€ ≤ 80 π 80 90 60 80 < π€ ≤ 100 80 Weight (in kg) Step – 1: πΉ π. π. = π. π€. Step – 2: Bar 1: π. π. = Bar 2: π. π. = Bar 3: π. π. = Bar 4: π. π. = 80 10 90 15 60 5 80 20 =8 =6 = 12 =5 Formula List β Median: (a) Median of π, π, π, π, π = π₯ (b) Median of π, π, π, π, π, π = (c) = 2.5 π΄ππππ 0 1 2 3 π 3 10 12 15 π. π. 3 13 25 40 4 5 8 2 48 50 = π π΄ππππ 0 1 2 3 π 3 10 12 15 π. π. 3 13 25 40 4 5 8 3 48 51 = π π 0 < π₯ ≤ 10 10 < π₯ ≤ 12 12 < π₯ ≤ 20 π 5 8 10 π. π. 5 13 23 = π (d) (e) 2+3 2 π 2 = 50 2 = 25 Median = 25π‘β π‘πππ + 26π‘β π‘πππ 2 ∴ Median = π+1 2 = 51+1 2 2+3 2 = 2.5. = 26 ∴ Median = 26π‘β π‘πππ = 3. π 2 = 23 2 = 11.5. ∴ Median group = 10 < π₯ ≤ 12. β Mean: (a) Mean of π, π, π, π = (b) (c) π₯+π¦+π§+π€ 4 π 0 1 π 4 19 ππ 0 19 2 3 4 25 29 23 50 87 92 π 0 < π₯ ≤ 10 10 < π₯ ≤ 12 π 5 11 π 2 5 ππ 10 55 12 < π₯ ≤ 20 16 4 64 ∑ π = 100, ∑ ππ₯ = 248 ∴ Mean = ∑ ππ₯ ∑π = 2.48 ∑ π = 11, ∑ ππ₯ = 129 ∴ Mean = ∑ ππ₯ ∑π = 11.73 Formula List β Mode: (a) π, π, π, π, π, π, π → Mode = π¦ and π. (b) π, π, π, π, π, π, π → Mode = π and π. (c) π π (d) 0 < π₯ ≤ 10 10 < π₯ ≤ 12 12 < π₯ ≤ 20 0 2 1 20 2 25 3 25 15 10 2 4 16 5 3 Modal group → 0 < π₯ ≤ 10. β Bar Chart: 1. Aadil observed the number of people in each of 20 cars entering a car park. The results are shown in the table below. Number of people Frequency 1 2 2 7 3 6 (a) Draw a bar chart to represent this information. (b) Write down the mode. (c) Calculate the mean number of people in each car. Answer: (a) (b) Mode = 2. (c) Mean = (1×2)+(2×7)+(3×6)+(4×4)+(5×1) 2+7+6+4+1 = 2.75. 4 4 5 1 Mode = 2 and 3. Formula List β Pie Chart: 1. The proportion of drinks sold at the refreshment tent at a summer fair are represented by the pie chart in figure. Only four categories of drink shown in the pie chart were sold and the number of coffees sold was 135. Coffee (a) Calculate the total number of drinks sold. (b) Calculate the number of orange juice drinks sold. 75° 100° 180 lemonade drinks were sold. Lemonade (c) Calculate the angle for lemonade drink. Tea 45° 140° Orange (d) Calculate the number of teas sold. Answer: (a) 75° ≡ 135 135 ⇒ 360° ≡ × 360 75 ∴ Total drinks = 648. (c) 135 ≡ 75° 75 ⇒ 180 ≡ × 180° 135 ∴ Lemonade ∠ = 100°. (b) 75° ≡ 135 ⇒ 140° ≡ 252 ∴ Orange juice = 252. (d) Tea = 360° − 75° − 100° − 140° ∴ Tea ∠ = 45°. 75° ≡ 135 135 ⇒ 45° ≡ × 45° 75 ∴ Tea = 81. 2. (a) Three times as many students said Blue as said Green. Calculate the angle of the sector which represents the number of students who said Green. (b) There are 8 more students who said Red than Yellow. How many students said Red? Answer: (a) Let Green be π₯° ∴ Blue = 3π₯° ∴ 90° + 110° + π₯° + 3π₯° = 360° ⇒ π₯ = 40° ∴ Angle for green = 40°. Red Yellow (b) 20° ≡ 8 students 1° ≡ 8 20 110° ≡ students 8 20 × 110° = 44 students ∴ Red = 44 students. Green 110° Blue Formula List EQUATION GRAPH β Distance-Time Graph and Speed-Time Graph: Constant acceleration At rest Constant speed Constant speed (Returning) Constant deceleration Formula List • • • • • # Distance time: • Gradient = Speed • Vertical difference = Distance • Average speed = β Set of values of π: Total distance Total time 0 to π1 = constant acceleration π1 to π2 = constant speed π2 to π3 = constant deceleration π3 to π4 = at rest π4 to π5 = returning # Speed time: • Gradient = Acceleration • Area under the graph = Distance • Average speed = Total distance Total time Formula List β Gradient at given point: To find gradient at π = π. π: Step 1: Draw the tangent line at π₯ = 1.5 Step 2: Take two suitable values from the tangent π¦ −π¦ Step 3: ππ₯=1.5 = π₯2 −π₯1 2 = 1 0.3−(−4) 3−0.75 = 1.91 β Solve Graphically: To solve ππ − π − π = π: Graph Equation = Line βΉ 2π₯ 2 − 2π₯ − 2 = 0 [× 2] βΉ 2π₯ 2 − 2π₯ − (2π₯) − 2 + (1) = 0 − (2π₯) + (1) βΉ 2π₯ 2 − 4π₯ − 1 = −2π₯ + 1 βΉ π¦ = −2π₯ + 1 π π 0 1 2 −3 ∴ Solution: π₯ = 2, −0.6 β Common Sketching: