Pair of linear equation s in two variables and Circles

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ALTERNATIVE ACEDEMIC CALENDER [2021 – 22]

STANDARD: 10th MONTH: September SUBJECT: Mathematics

Sl No

Month/ Week

Expected learning

competencies

Suggested Learning Activities Evaluation

01

Sep

tem

ber

firs

t w

eek (

Th

ree p

eri

od

s)

Defining

Equation, Pair

of linear

equations in

two variables

and Pair of

linear equation

in one variable.

Students already know about

Equation, Linear equation in

one variable, when we teach

Linear equation in two

variables, then they are able

to differentiate between

Linear equation in one

variable and Linear equation

in two variables.

Using Activity

sheet – 1.

02 Graphical

representation

of Pair of linear

equations in

two variables

and Pair of

linear equation

in one variable.

Explain how we can draw a

pair of linear equation in one

variable on Number line and

also graphically represent

solution of pair of linear

equations in two variables as

two lines.

Using Activity

sheet – 2.

03 Defining and

comparing the

ratios of linear

equations in

two variables.

Practicing to find out ratios of

coefficients and comparing

the ratios of linear equations

in two variables.

https://www.youtube.com/watch

?v=nRGJruWqXwU

Using Activity

sheet – 3.

Solving

problems in

Work Book –

part1, Page No-

21 & 22, Main I,

II and III

Pair of linear equations in two variables and Circles

2 | P a g e

04

Sep

tem

ber

seco

nd

week(

two

peri

od

s)

Graphical

Method of

solution of a

Pair of Linear

Equations.

Practicing to find solution of a

Pair of Linear Equation in two

variables by Graphical

Method and also steps to

follow to find solution


?v=J88VLDDQKgI

Using Activity

sheet – 4.

Solving

problems in

Work Book –

part1, Page No-

23 and Page

No- 24 - Main

I(a).

05 Algebraic

Methods of

solving a Pair

of Linear

equations:

1st method:

Substitution

method

Practicing the Definition and

step to find solutions of Pair

of Linear Equation in two

variables by using

Substitution method.

Using Activity

sheet – 5.

Solving

problems in

Work Book –

part1, Page No-

23and page no -

24 - Main I(b).

06

Sep

tem

ber

thir

d w

eek(

thre

e p

eri

od

s)

Algebraic

Methods of

solving a Pair

of Linear

equations:

2nd method:

Elimination

method




variables by using Elimination

method.


?v=ivlTc6JOhuc

https://www.youtube.com/wat

ch?v=dbf5_jdTsa8

Using Activity

sheet – 6.

Solving

problems in

Work Book –

part1, Page No-

23and page no -

25 - Main I(c).

07 Algebraic

Methods of

solving a Pair

of Linear

equations:

3rd method:

Cross

Multiplication

Method




variables by using Cross

Multiplication method.

Using Activity

sheet – 7.

Solving

problems in

Work Book –

part1, Page No-

23and page no -

25 - Main I(d).

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08 Equations

reducible to a

pair of linear

equations in

two variables.

Discussion to solve such

pairs of equations which are

not linear but can be reduced

to linear pair form making

some suitable methods like

Graphical method or any one

of Algebraic method to solve

pair of equations.


ch?v=Kp_oKu5KW_Y

Using Activity

sheet – 8.

09

Sep

tem

ber

4th

week (

Fo

ur

peri

od

s)

Circle and a

line

Defining Circles and

explaining about relationship

between circle and a line.

Using Activity

sheet – 9.

10 Chapter:

Circles

Theorem – 01

(Theorem on

tangents)

Practicing the steps involved

in proving theorem – The

tangent at any point of a

circle is perpendicular to the

radius through the point of

contact.

https://www.youtube.com/wa

tch?v=IYWfUrtDJBo

https://www.youtube.com/wa

tch?v=Lz6i-Wep5wM

Using Activity

sheet – 10.

Solving

problems in

Work Book –

part1, Page No-

38– Practicing

theorem 01

11 Chapter:

Circles

Theorem –

02(External

point theorem)

Practicing the steps involved

in proving theorem – The

lengths of tangents drawn

from an external point to a

circle are equal.


ch?v=y2_4CoDboJs

Using Activity

sheet – 11.

Solving

problems in

Work Book –

part1, Page No-

39– Practicing

theorem - 02

12 Problems

related to

theorems.

Practicing problems related to

theorems on circles.

Using Activity

sheet – 12.

Equation:An equation is a mathematical sentence/statement that has two

equal sides separated by an equal sign.

Example: 4 + 6 = 10 is an example of an equation. We can see on the l

of the equal sign, 4 + 6 and on the right hand side of the equal sign 10.

Linear equation in one variable:

equation which is expressed in the form of ax + b = 0

integers, and x is a variable and has only one solution.

Example:5x + 10 = 0 is a linear equation having a single variable in it.

x is – 2.

1) If 6y – 15 = 0, then find the value

of y. Solution: 6y – 15 = 0

6y = 15

y = ��

= ��

3) If 5t – 50 = 0, then find the value

of t. Solution:

Linear equations in two variables

An equation in the form of ax+by+c

a, b are not equal to zero is called a linear equation

power of variable is 1 and it has infinitely many solutions.

Example: In x + y – 5 = 0 equation, the solution is

1) Find one value of x and y which satisfies the equation 2x + y = 7.

Pair of Linear Equations in Two

An equation is a mathematical sentence/statement that has two

equal sides separated by an equal sign.

is an example of an equation. We can see on the l


Linear equation in one variable:The linear equations in one variable is an

is expressed in the form of ax + b = 0, where a and b are two

variable and has only one solution.

5x + 10 = 0 is a linear equation having a single variable in it.

15 = 0, then find the value 2) If 2y + 8 = 0, then find the value

of y. Solution:

50 = 0, then find the value 4) If 3t + 19 = 0, then find the value

of t. Solution:

in two variables:

An equation in the form of ax+by+c = 0, where a,b and c are real num

called a linear equations in two variables.

power of variable is 1 and it has infinitely many solutions.

= 0 equation, the solution is x = 2 and y = 3.

Find one value of x and y satisfies the equation

2) Find one value of x which satisfies the equation x = 4y.

Pair of Linear Equations in Two Variables ACTIVITY

An equation is a mathematical sentence/statement that has two

is an example of an equation. We can see on the left side


The linear equations in one variable is an

, where a and b are two

5x + 10 = 0 is a linear equation having a single variable in it. Value of

ind the value

If 3t + 19 = 0, then find the value

, where a,b and c are real numbers and

in two variables. Highest

x = 2 and y = 3.

Find one value of x and y which satisfies the equation

ACTIVITY - 1

Graphical representation of linear equation in one variable:

Graphical representation of linear equation in one variable is actually locatina value on a number line.

For example:2k + 10 = 0, then

Graphical representation of linear equation

Graphical representation of linear equation in two variables is actually a straight line. It may be Horizontal or vertical line.

For example:x – 2y = - 5 then graphical representation is given below.

If y = 0 then x = - 5, y = 1 then x =

When 2x + y = 6 then find the solution graphically.

Pair of Linear Equation

If 6y – 18 = 0 then represent y value in graph sheet

Graphical representation of linear equation in one variable:

Graphical representation of linear equation in one variable is actually locatin

2k + 10 = 0, then graphical representation is given below.

Graphical representation of linear equations in two variables:

Graphical representation of linear equation in two variables is actually a line. It may be Horizontal or vertical line.

5 then graphical representation is given below.

5, y = 1 then x = - 3

When 2x + y = 6 then find the solution graphically.

ACTIVITY Pair of Linear Equations in Two Variables

present y value in graph sheet

Graphical representation of linear equation in one variable is actually locating

representation is given below.

Graphical representation of linear equation in two variables is actually a

5 then graphical representation is given below.

ACTIVITY - 2

Pair of Linear Equations in Two Va

The general representation of a pair of linear equation

and y is a1x + b1y + c1 = 0 and

real numbers and a12 + b1

2 ≠ 0, a

For example:10x + 4y = 3 and

variables.

Comparing the ratios of Pair of Linear Equations in Two Variables:

Compare the

ratios

��

��

��

Number of solutions

Exactly one solution

Types of solution

Consistent and Independent

Graphical representation

Intersecting lines

x – 2y = 0 and

3x + 4y = 20

Comparing the ratios

Number of solutions

Types of solution

Pair of Linear Equation

Pair of Linear Equations in Two Variables:

general representation of a pair of linear equations in two variables

= 0 and a2x + b2y + c2 = 0 where a1, b1, c1, a

≠ 0, a22+ b2

2 ≠ 0.

10x + 4y = 3 and – x + 5y = 2 are Pair of linear equations in two


��

��

��

��

��

��

�

�

��

Exactly one solution

Infinitely many solutions

No solution

Consistent and Independent

Consistent and Dependent

Inconsistent

Intersecting lines Coincident lines

Parallel lines

2y = 0 and 3x + 4y = 20

2x + 3y = 9 and 4x + 6y = 18

x + 2y = 4 and2x + 4y = 12

ACTIVITY

Pair of Linear Equations in Two Variables

in two variables say x

, a2, b2, c2 are all

x + 5y = 2 are Pair of linear equations in two


��

��

��

��

�

�

No solution

Inconsistent

Parallel lines

x + 2y = 4 and 2x + 4y = 12

ACTIVITY - 3

Graphical method of solution of a pair of

linear equations in two variables:

To find a solution of a pair of linear equation

in two variables, we substitute assumed value

of one variable to another variable

1) 10 students of class X took part in

four more than the number of boys, then find the number of boys and the number

of girls who took part in the quiz.

Total number of students who took part in

Let, the number of boys isxand, number

According to question,

x + y =10 ...........(1)

x – y = 4 ..........(2) From equation (1)

From equation (2)

Therefore x = 7 and y = 3

2) The cost of 5 pencils and 7 pens is

₹46. Find the cost of each pencil and each pen.

Let the cost of each pencil be x &the cost of each pen be y From equation (1)

From equation (2)

Therefore x = and y =

x 5 4 6

y=10-x 5 6 4

x 5 4 3

y = x - 4 1 0 -1

x

y

x

y


Graphical method of solution of a pair of

in two variables:

To find a solution of a pair of linear equations

, we substitute assumed value

of one variable to another variable

took part in a mathematics quiz. If the number of girls is


of girls who took part in the quiz.

Total number of students who took part in the quiz =10

and, number of girls who took part in the quiz

cost of 5 pencils and 7 pens is ₹50. The cost of 7 pencils and 5 pens is

46. Find the cost of each pencil and each pen.

Let the cost of each pencil be x &


Graphical

a mathematics quiz. If the number of girls is


of girls who took part in the quiz =y

The cost of 7 pencils and 5 pens is

ACTIVITY - 4

Graphical

Method

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Algebraic methods of solving a pair of linear equations in two variables:

Substitution Method:We have to substitute the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. So this method is known as the substitution method. For example:x + y = 14 and x – y = 4 Step - 1:Find the value of one variable, say y in terms of the other variable. x from either equation, Whichever is convenient. x + y = 14 ……(1) and x – y = 4 …….(2) , from equation (1) y = 14 – x Step - 2:Substitute this value of y in the other equation and reduce it to an equation in one variable that is in terms of x, which can be solved. Substitute the value of y = 14 – x in (2),x – 14 + x = 4 ⇒ 2x = 18 ⇒ x = 9 Step - 3:Substitute the value of x (or y) obtained in step 2 in the equation used in step 1 to obtain the value of the other variable. Substitute x = 9 in equation (1), x + y = 14 ⇒ 9 + y = 14 ⇒ y = 14 – 9 = 5 Solve the following pair of linear equation by the substitution method:

1) 2x + 3y = 11 and 2x - 4y = - 24 Solution:

2) 3x - y = 3 and 9x - 3y = 9 Solution:

3) x + 2y = 4 and 2x + 4y = 12 Solution:

4) 2x + 3y = 9 and 4x + 6y = 18 Solution:

5) s - 7t = - 4 and s - 3t = 6 Solution:

6) 2x + 3y = 13 and 4x + 5y =23 Solution:

ACTIVITY - 5 Pair of Linear Equations in Two Variables

Substitution

method

9 | P a g e


Elimination Method:

In the elimination method, we eliminate any

one of the variables by using basic arithmetic

operations and then simplify the equation to

find the value of the other variable. Then we

can put that value in any of the equations to

find the value of the variable eliminated.

For example:x + y = 5 and 2x - 3y = 4.

Step 1:Let x + y = 5 →(1) and 2x - 3y = 4 →(2).

For equating x coefficient, multiply equation (1) by 2

We get, 2x + 2y = 10 → (3) and 2x - 3y = 4 → (2).

Step 2:Subtract equation (3) from (2), Now we get, -5y = - 6 ⟹ 5y = 6 ⟹⟹⟹⟹ y =��

Step 3:Substituting y value in equation (1), x + y = 5 ⟹⟹⟹⟹ x +��= 5 ⟹⟹⟹⟹ x =

��

.

Solve the following pair of linear equation by the elimination method:

1) 2x + 3y = 8 and 4x + 6y = 7 Solution:

2) x + y = 6 and y – x = 2 Solution:

3) x + y = 5 and 2x - 5y = 4 Solution:

4) 3x + 4y = 10 and 2x - 2y = 2 Solution:

5) 3x - 5y = 4 and 9x - 2y = 7 Solution:

6) 3x + 4y = - 6 and 3x – y = 9 Solution:

ACTIVITY - 6 Pair of Linear Equations in Two Variables

Elimination

Method

Algebraic methods of solving a pair of linear equation

The arrows between the two numbers in

and the second product is to be subtracted from the first.

For example:2x + 3y = 46 and 3x + 5y = 74

Step-1:Write the given equation in the form

c2 = 0, then we get 2x + 3y – 46 = 0 a

Step-2:Taking the help of the diagram, write the equation in the form �

�� =

��

= �

��

��

= �

�� =

�

�� =

Step-3: When x and y, provided

Solve the following pair of linear equation having unique solution by using

cross multiplication method:

1)x - 3y – 3 = 0 and 3x - 9y – Solution:

3) 3x - 5y = 20 and 6x - 10y = 40Solution:


Cross

multiplication

method


The arrows between the two numbers indicate that they are to be multiplied

and the second product is to be subtracted from the first.

2x + 3y = 46 and 3x + 5y = 74

Write the given equation in the form a1x + b1y + c1 = 0 and a

46 = 0 and 3x + 5y – 74 = 0.

Taking the help of the diagram, write the equation in the form

= �

� =

�

�� =

�

�

: When x and y, provided a1 b2 - a2 b1≠ 0. Here values of x = 8 and y = 10.


2 = 0 2) 2x + y = 5 and 3x + 2y = 8Solution:

10y = 40 4) x - 3y – 7 = 0 and 3x Solution:


in two variables:

dicate that they are to be multiplied

= 0 and a2x + b2y +

Taking the help of the diagram, write the equation in the form

0. Here values of x = 8 and y = 10.


2x + y = 5 and 3x + 2y = 8

7 = 0 and 3x - 3y – 15 = 0

ACTIVITY - 7

Multiply equation (1) by 2 and equation (2)

elimination method we get, m = 2 and n = 3

Now �

� = 2 and

��

= 3, OR x =

Solve the following pairs of equations by reducing them to a pair of linear equations:

��

√� +

�

�� = 2 and

�

√� � �

�� = -1.

3) �

� � � +

��

= 2 and �

� � � -

�


Equations reducible to a pair of linear equations in two variables:

The solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable methods like graphical or Algebraic methods.

Example: �

��+

��

= 2 and�

��+

��

=��

Let ��= mand

��

= n

Now, ��

+ ��= 2 ⇒⇒⇒⇒ 3m + 2n = 12

and��

+ ��=

��

⇒⇒⇒⇒ 2m + 3n = 13⟶

Multiply equation (1) by 2 and equation (2) by 3. By simplification using

elimination method we get, m = 2 and n = 3

= 3, OR x = ��and y =

��.

Solve the following pairs of equations by reducing them to a pair of linear

1. �� + 3y = 14 and

��- 4y = 23.

� � �

= 1 4) ��

� � � +

��

= 4 and ��

� �


Equations reducible to a pair of linear equations

of such pairs of equations which are not linear but can be reduced to linear form by making some suitable methods like

3m + 2n = 12 ⟶ (1)

⟶(2)

By simplification using

Solve the following pairs of equations by reducing them to a pair of linear

4y = 23.

��

� � -

�

� � � = -2

ACTIVITY - 8

Circle and a line:

Circle:

A circle is all points in the same plane

that lie at an equal distance from a

centre point.

Radius:

Radius of a circle is the distance from

thecentre of the circle to any point

onit’scircumference.

Diameter:

The diameter of a circle is a line

segment that passes through the centre

of a circle and has two endpoints at the

circumference.

Chord:

A chord of a circle is a straight line segment whose endpoints both lie on a

circular arc.

Tangent:

A tangent to a circle is a straight line which

point.P - point is called the point of tangency. The tangent

perpendicular to the radius at the point of tangency.

Secant:

A secant of a circle is a line that intersects a circle at two distinct points

Secant of a circle.

Fill in the blanks with appropriate answers:

1) A tangent to a circle intersects it in ______________ points.

2) _____________ number of tangents can be drawn to a circle passing

through a point lying inside the circle.

3) A circle can have ______________

points of diameter.

4) The common point of a tangent to a circle and the circle is called

_____________________.

5) The angle between tangent to a circle and radius drawn at a point of co

is ___________.

CIRCLES

A circle is all points in the same plane

that lie at an equal distance from a

Radius of a circle is the distance from

of the circle to any point

line

segment that passes through the centre

of a circle and has two endpoints at the

is a straight line segment whose endpoints both lie on a

a straight line which touches the circle at only one

point is called the point of tangency. The tangent XY to a circle is

perpendicular to the radius at the point of tangency.

a line that intersects a circle at two distinct points

Fill in the blanks with appropriate answers:

tangent to a circle intersects it in ______________ points.

_____________ number of tangents can be drawn to a circle passing

through a point lying inside the circle.

an have ______________ parallel tangents at the most at the


between tangent to a circle and radius drawn at a point of co

ACTIVITY CIRCLES

is a straight line segment whose endpoints both lie on a

touches the circle at only one

to a circle is

a line that intersects a circle at two distinct points. AB is

_____________ number of tangents can be drawn to a circle passing

parallel tangents at the most at the end


between tangent to a circle and radius drawn at a point of contact

ACTIVITY - 9

Theorem

The tangents at any point of a circle are

the point of contact.

Data:

To prove:OP⊥XY

Construction: Take any point on Q, other

than P on the tangent XY and join OQ.

It meets circle at a point R.

Proof:

Statement

Hence, Q is a point on the tangent XY, other than the point of contact P. Sthe circle.

Let, OQ intersect the circle at R OP=OR

Now, OQ = OR + RQ

OQ >OR

OQ > OP

__________ is the shortest distance to the tangent from the centre ‘O’

OP ⊥⊥⊥⊥XY

A tangent PQ at a point P of a circle of radius

centre O at a point Q so that OQ

According to Pythagoras theorem, OQ2 = PQ2+ OP2

OP2=144-25 =119

OP=√�� cm.

CIRCLES

You Must Know:

There is one and only one tangent to a circle passing

through a point lying on the circle.

The line containing the radius through the point of

contact is also the normal to the circle at that point.

Theorem – 01[Theorem on tangents]

tangents at any point of a circle are perpendicular to the radius through

: Take any point on Q, other

than P on the tangent XY and join OQ.

Statement Reasons

Hence, Q is a point on the tangent XY, other than the point of contact P. So Q lies outside

[There is only one point of a contact to a tangent ]

Let, OQ intersect the circle at R OP=OR ( )

( )

( )

( )

__________ is the shortest distance to the

( )

of a circle of radius 5cm meets a line through the

OQ=12cm. Find the length of PQ.

According to Pythagoras theorem,

CIRCLES ACTIVITY

There is one and only one tangent to a circle passing

through a point lying on the circle.

The line containing the radius through the point of

contact is also the normal to the circle at that point.

perpendicular to the radius through

Reasons

ly one point of a contact to a tangent ]

( )

( )

( )

( )

( )

meets a line through the

ACTIVITY - 10

Theorem – 02

The lengths of tangents drawn from an external point to a circle are equal.

Data:

To prove:PB=BQ

Construction:

Proof:

Statement

In ∆ APB &∆ AQB

∠APB= ∠AQB=900

AB=AB

AP=AQ

∴ ∆ APB ≅ ∆AQB

BP=BQ.

The length of a tangent from a point A at distance

is 4 cm. Find the radius of the circle.

Solution:From Pythagoras theorem,

AO2 = AB2 + OB2

OB2 = 25 – 16 = 9 OB = 3 cm

CIRCLES

You Must Know:

There is no tangent to a c

point lying inside the circle.

There is one and only one tangent to a circle

passing through a point lying on the circle.

There are exactly two tangents to a circle through a

point lying outside the circle.


Reasons

The length of a tangent from a point A at distance 5 cm from the centre of circle

cm. Find the radius of the circle.

From Pythagoras theorem,

16 = 9 OB = 3 cm

CIRCLES

ACTIVITY

Know:

There is no tangent to a circle passing through a

point lying inside the circle.




point lying outside the circle.


cm from the centre of circle

ACTIVITY - 11

ircle passing through a




1) From the point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is: A) 7 cm B) 12 cm C) 15 cm D) 24.5 cmSolution:

2) In the given figure, if TP and TQ are tangents to a circle with centre ‘O’

so that

∠∠∠∠POQ = 1100, then ∠∠∠∠PTQ is equal to:

A) 600 B) 700

C) 800D)900

Solution:

3) If tangents PA and PB from a point P to a circle with centre O are inclined to

each other at angle of 800, then A) 500 B) 600

Solution:

4) A tangent PQ at a point P of a circle of radius 5 cm meets a line through the

centre O at a point Q, so that A) 22 cm B) 23 cm Solution:

5) A tangent PQ at a point P of a circle of radius 6 cm meets a line through

centre ‘O’, so that OQ = 10 cm then the length of PQ is . . . . .

A) 9 cm B) 8 cm

Solution:

CIRCLES

m the point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is:

A) 7 cm B) 12 cm C) 15 cm D) 24.5 cm

TP and TQ are tangents to a circle with centre ‘O’

PTQ is equal to:

If tangents PA and PB from a point P to a circle with centre O are inclined to

, then ∠∠∠∠POA is equal to: C) 700 D) 800

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the

OQ=12 cm. Length of PQ is : C) 24 cm D) 25 cm

A tangent PQ at a point P of a circle of radius 6 cm meets a line through

centre ‘O’, so that OQ = 10 cm then the length of PQ is . . . . .

C) 7 cm D) 6 cm

ACTIVITY CIRCLES

m the point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is:

A) 7 cm B) 12 cm C) 15 cm D) 24.5 cm

TP and TQ are tangents to a circle with centre ‘O’

If tangents PA and PB from a point P to a circle with centre O are inclined to

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the

25 cm

A tangent PQ at a point P of a circle of radius 6 cm meets a line through

ACTIVITY - 12

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Pair of linear equation s in two variables and Circles

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