Linear Equation

• The general form of linear equation in two variables -
  ax + by + c = 0
  note: x & y are variables and
  a,b,c are numerical coefficient
• The solution of linear equation is a pair of values x & y which will satisfy the equation
  eg. 2x + y = 5
  x = 2 & y = 1
• The graph of linear equation in two variable is a line.
  Every point on the graph represents a solution of the equation.
  The graph represents an equation in two variables.
  

Pair of Linear Equation
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Each equation is represnted by a line. So there will be Two Lines.
This two lines can be in 3 ways

• Intersecting Lines
• Parallel Lines
• Coincident Lines

Condition for solution

• a1		b1	Unique Solution      Intersecting Lines
  --	≠	--
  a2		b2

• a1		b1		c1	No Solution      Parallel Lines
  --	=	--	≠	--
  a2		b2		c2

• a1		b1		c1	Infinitely many Solution      Coincident Lines
  --	=	--	=	--
  a2		b2		c2

General Form of Quadratic Equation
ax² + bx + c = 0
where x is variable abd a,b,c are real numbers & a is not equal to 0

Roots of Quadratic Equation = Zeros of a Quadratic Polynominial.

There are 2 roots for a given Quadratic Equation

There are three methods to find the roots of Quadratic Equation

• Factorisation Method
• Using Quadratic Formula

Method II - Using Quadratic Formula to find the roots of Quadratic Equation

ax² + bx + c = 0



Discriminant = b² - 4ac

Discriminant determines the Nature of Roots

Discriminant	Nature of Roots
+ve	real and distinct
0	real and equal / same / repeated
-ve	not real

Word Problems based on age

• Step1 - Create the quadratic equation from given age data
• Step2 - Find roots the quadratic equation

Word Problems based on speed distance & time

• Step1 - Create the quadratic equation from given age data
• Step2 - Find roots the quadratic equation

Word Problems based on work & time

• Step1 - Create the quadratic equation from given work & time data
• Step2 - Find roots the quadratic equation

Arithmetic Progression(AP):-is a list of number in which each term is obtained by adding a fixed number to the preceding term, except the first term

Fixed number: Common difference (d)
Term: Number in the sequence (a1, a2, a3)

• First Term --> a1 or a
• Second Term --> a2
• Third Term --> a3
• nth Term --> an
• Fixed number: Common difference (d)

General Form of an AP is
a, a+d, a+2d, a+3d....

Formula 1: a_n = a + (n-1)d Note:

• An AP can be infinire
• An AP can be finite last term is an is l
• 'd' can be positive , 0, negative

Formula 1: a_n = a + (n-1)d

a_n --> nth term
a --> 1^st term
n --> number of term
d --> Common Difference

Type I Question to Find a_n or

From above 3 values are given and asked to find the 4^th one

Find the 20 Term from Last Term
Note:- Reverse the AP and then fidn 20 Term from first

Type II: To Find 'n'
ex. Which term of the AP 3,8, 13, 18..... is 78 ?

Formula 2:Sum of first n terms of am AP

Sn =	n [ 2a + (n-1)d ]	Given a, d, n
	2
Sn =	n [ a + l ]	Given a, l=an, n
	2

Similar triangle condition

Example based on similar triangles.

Basic Proportionality Theorem (BPT) Defination & examples

Converse of BPT Defination

Criteria for Similar Triangle - AAA / AA / SAS / SSS

Proffs Based on Criteria for Similar Triangles

Second theorem - Theorem of Area of Similar Triangle Defination & examples

Objective Questions based on Pythagoras Theorem

Problems on Triangles

Problems on Triangles

Problems on Triangles Prove That Type

Problems on - Theorem of Area of Similar Triangle

Problems on - Pythagoras Theorem

Two Steps Problems to Prove L.H.S = R.H.S

Two Steps Problems to Prove L.H.S = R.H.S

Three Steps Problems to Prove L.H.S = R.H.S

Rider From Board Paper 2012 & 2013

Rider From Board Paper 2014

Rider From Board Paper 2015

Rider From Board Paper 2016 & 2018

Rider From Board Paper 2019

Rider From Board Paper 2020

Rider From Board Paper 2010 & 2011

Rider From Prelims Paper 2016 & 2020

Rider From Prelims Paper 2019 & 2017

Revision 1

Revision 2

Formula 1 - Distance Formula



Type 1 - Distance between points

• Find distance between points A(x1, y1) & B(x2, y2)

Type 2 - About geometric figures

• Check weather A(x1,y1), B(x2,y2) & C(x3.,y3) are vertices of an isosceles or right angle triangle
• Name the type of quadrilateral if any formed by points A(x1,y1), B(x2,y2) ,C(x3,y3) & D(x4,y4)
  
  

  Quadrilateral	Sides	Diagonals
  Square	All sides are equal	Equal
  Rhombus	All sides are equal	Not Equal
  Rectangle	Opp sides are equal	Equal
  Parallelogram	Opp sides are equal	Not Equal

Type 3 - Question about a point at equal distance from other two points

• If Q(0,1) is equi distance from P(5,-3) and R(x,6) fidn the values of x
• Find the coordinates of a Point on x-axis, which is equidistance from A(2,-5) and B(-2,9)
• Find a relation between x and y such that the point P(x,y) is equi-distance from Q(3,6) and R(-3,4)

Note:-

if a point P lies on X-axiz then its y coordinate is 0 eg.P(x,0)
if a point P lies on Y-axiz then its x coordinate is 0 eg.P(0,y)
Type 2 - About geometric figures

• Check weather A(x1,y1), B(x2,y2) & C(x3.,y3) are vertices of an isosceles or right angle triangle
• Name the type of quadrilateral if any formed by points A(x1,y1), B(x2,y2) ,C(x3,y3) & D(x4,y4)
  
  

  Quadrilateral	Sides	Diagonals
  Square	All sides are equal	Equal
  Rhombus	All sides are equal	Not Equal
  Rectangle	Opp sides are equal	Equal
  Parallelogram	Opp sides are equal	Not Equal

Type 4 - To show that the given points are collinear

• Given three points A(1,2), B(5,3) & C(18,6). Determine weather they are collinear
  
  Solution
  
  
  • Steps 1 - Find Distance Between First two points AB
  • Steps 2 - Find Distance Between Last two points BC
  • Steps 3 - Find Distance Between First point & Last point AC
  • Steps 4 - Adding any two distances if we get the third distance then 3 points are collinear

Formula 2 - Section Formula



point P (x,y) divides line segment joining A(x₁,y₁) and B(x₂,y₂) in the ration m1 : m2

x	=	m1x2 + m2x1		y	=	m1y2 + m2y1
		m1 + m2				m1 + m2

Formula 3 - MidPoint Formula



point P (x,y) is the mid point of line segment joining A(x₁,y₁) and B(x₂,y₂) in the ration m1 : m2

x	=	x1 + x2		y	=	y1 + y2
		2				2

Type1 - To find the coordinates of the point of division

Type 2 - To find the ration when all the points are given

Type 4 - Problems based on geometrical figure

Ex. Show Points A(1,0) B(5,3) C(3, 7) D)-2,4) are vertices of parallelgram

Solution - Diagional of parallelgram bisect each other

• Step 1 - Find the Mid Point Of Diagonal AC using mid point formula
• Step 2 - Find the Mid Point Of Diagonal BD using mid point formula
• From Step 1 & Step 2 - if the midpoint cordinates of both Diagonal are same then Diagional Bisect each other.
  Therefor Quadrilateral ABCD is a parallelgram

Formula 4 - Area of Triangle

ar(ABC)	=	1	[ x1(y2-y3) + x2(y3-y1) + x3(y1-y2) ]
		2

Type1 - To find area when the vertices are given

Type3 - If are of triangle is given then find one unknown coordinate

Type2: To find the are of a quadrilateral where coordinates of all 4 vertices are given

Type4: To find the value of unknown coordinates, if the given three points are collinear

Trignometric ratios of angles
sinΘ cosΘ tanΘ cosecΘ secΘ cotΘ

Questions based on Trignometric ratios

Relation of trignometric ratios amoung each other
 

• Reciprocals
   

  • sinΘ =

    1 cosecΘ

  • cosΘ =

    1 secΘ

  • cotΘ =

    1 tanΘ

• Relation between sin cos & tan
   

  • tanΘ =

    sinΘ cosΘ

  • cotΘ =

    cosΘ sinΘ

• Relation between complementary angles in right angle triangle
   
  • sinΘ = cos(90 - Θ)
  • tanΘ = cot(90 - Θ)
  • secΘ = cosec(90 - Θ)

Trignometric ratios of specific angles

Angle	0	30	45	60	90	Note
Sin	0	1 2	1 √2	√3 2	1	--
Cos	1	√3 2	1 √2	1 2	0	reverse of sin
Tan	0	1 √3	1	√3	n.d	sin/cos
Cosec	n.d	2	√2	2 √3	1	1 / sin
Sec	0	2 √3	√2	2	n.d	1 / cos
Cot	n.d	√3	1	1 √3	0	cos/sin

Questions based on Trignometric ratios of specific angles

Elementary Trigonometric Identies To Prove below

• sin²Θ + cos²Θ = 1
• 1 + tan²Θ = sec²Θ
• 1 + cot²Θ = cosec²Θ

To Prove Trignometric Identities

Points to Remember

• Convert the rations in terms of sin and cos
   

  • tanΘ =

    sinΘ cosΘ

  • cotΘ =

    cosΘ sinΘ

  • secΘ =

    1 cosΘ

  • cosecΘ =

    1 sinΘ

• All methods of factorisation and expression
   
  • common factor
  • factorisation by grouping
  • a² + 2ab + b² = (a+b)²
  • a² - 2ab + b² = (a-b)²
  • a³ + b³ = (a+b)(a²-ab+b²)
  • a³ - b³ = (a-b)(a²+ab+b²)
  • (a-b)(a+b) = a²-b²
• Multiple by conjugate
   

  • 1              (1+sinΘ)

    *

    (1-sinΘ) (1-sinΘ)

  • 1              (1-cosΘ)

    *

    (1+cosΘ) (1+cosΘ)

• Use Elementary Identities
   
  • sin²Θ + cos²Θ = 1
  • sin²Θ = 1 - cos²Θ
  • cos²Θ = 1 - sin²Θ
  • --
  • 1 + tan²Θ = sec²Θ
  • tan²Θ = sec²Θ - 1
  • 1 = sec²Θ - tan²Θ
  • --
  • 1 + cot²Θ = cosec²Θ
  • cot²Θ = cosec²Θ - 1
  • 1 = cosec²Θ - cot²Θ

Class Intervals

• 1 - Continuous
• 2 - Non-Continuous

• Uppler limit & Lower limit
• Class mark = (Lower limit + Upper limit)/2
• Class size = Upper limit - Lower limit
• Frequency = fi
• Mean (Average) = Total_of_all_observations/no_of_observations
• Mean of grouped data

Find the Mean using Direct Method

Find the Mean using Assumed Method

Find the Mean using Step Deviation Method

Find the Mode of grouped data

Find the Median of grouped data

Test Exam

1) Probality - Theoretical
2) P(E) = No of outcome favourable to E / No of Possible outcome of the experiment
3) Elementary Event - An Event with one outcome of the experiement
4) Sum of probalities of all elementary events of an experiment is 1 (one)
5) Complementary Events
6) Impossible Events And Sure Evetns
7) 0 <= P(E) <= 1

Model paper sent by the boarda

Solve the following word problems on quadratic equations

Model paper for second term exam

Homework for today

Term2 Prelim Maths Paper