Linear Equations In Two Variables

 Mind Maps

Class IX - Maths: Linear Equations in Two Variables
Q) Write the equation 2x+3y=4 in the form ax+by+c=0 and indicate the value of a,b and c?

Q) Define Graphing Linear Equation?

Q) Solve the equation 3x - 1 = 15 - x and represent the solution graphically?

Q) Describe Cross Multiplication?

Q) The sum of two number is 16 and their difference is 4. Find the numbers?

Q) Write the following y=2 in two variable format?

Q) Find two solutions for the following equation 3y+4=0?

Q) What are the Quadrants in the Coordinate Plane?

Q) How to determine the coordinates of a point in a plane?

Q) Draw the graph of the equation 5x - y = 1?

Q) In a two digit number. The units digit is thrice the tens digit. If 49 is added to the number, the digits interchange their place.Find the number?

Q) Solve the equation 6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3)?

Contents
• Linear Equations
• Solvability of Linear Simultaneous Equations
• Word Problems on Simultaneous Linear Equations
• Solution Of a Linear Equations
• Cross Multiplication
• Coordinate Graph
• Coordinates of a Point
• Example to plot points on a coordinate graph paper
• Method to draw the graph of linear equation in two variables
• What is cross multiplication?
The process of multiplying the numerator on the left hand side with the denominator on the right hand side and
multiplying the denominator on left hand side with the numerator on right hand side is called cross multiplication.
(mx + n)/(ox + p) = q/r where m, n, o, p, q, r are numbers and ox + p ≠ 0
⇒ r(mx + n) = q(ox + p)

• Example:a_1x + b_1y + c_1 = 0 --------- (i)
a_2x + b_2y + c_2 = 0 --------- (ii)
we get: x/(b_1 c_2 - b_2 c_1) = y/(a_2 c_1 - a_1 c_2) = 1/(a_1 b_2 - a_2 b_1)
• Linear equation in two variables is represented graphically by the line whose points give the collection of solutions of the equation. This is called graphing linear equation.
• An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
• A linear equation in two variables has infinitely many solutions.
• The graph of every linear equation in two variables is a straight line.
• x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.
• The graph of x = a is a straight line parallel to the y-axis.
• The graph of y = a is a straight line parallel to the x-axis.
• An equation of the type y = mx represents a line passing through the origin.
• Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.
Linear Equations:

• Linear Equations Let us first recall what you have studied so far. Consider the following equation:
• 2x + 5 = 0 It
• solution, i.e., the root of the equation, is- 5 /2 . This can be represented on the number line as shown below: Fig.
• To remember the process of framing simultaneous linear equations from mathematical problems
• To remember how to solve simultaneous equations by the method of comparison and method of elimination
• To acquire the ability to solve simultaneous equations by the method of substitution and method of cross-multiplication
• To know the condition for a pair of linear equations to become simultaneous equations
• To acquire the ability to solve mathematical problems framing simultaneous equations
• We know that if a pair of definite values of two unknown quantities satisfies simultaneously two distinct linear equations in two variables,
then those two equations are called simultaneous equations in two variables.
We also know the method of framing simultaneous equations and two methods of solving these simultaneous equations.
• We have already learnt that linear equation in two variable x and y is in the form ax + by + c = 0.
• • Where a, b, c are constant (real number) and at least one of a and b is non-zero.
The graph of linear equation ax + by + c = 0 is always a straight line.
• Every linear equation in two variables has an infinite number of solutions.
Here, we will learn about two linear equations in 2 variables.
(Both equations having to same variable i.e., x, y).

Solvability of Linear Simultaneous Equations

• In the method of cross-multiplication, for the simultaneous equations,
a_1x + b_1y + c_1 = 0 --------- (i)
a_2x + b_2y + c_2 = 0 --------- (ii)
we get: x/(b_1 c_2 - b_2 c_1) = y/(a_2 c_1 - a_1 c_2) = 1/(a_1 b_2 - a_2 b_1)
that is,
• x = (b_1 c_2 - b_2 c_1)/(a_1 b_2- a_2 b_1) , y = (a_2 c_1 - a_1c_2)/(a_1 b_2 - a_2 b_1) --------- (iii) Now, let us see when the solvability of linear simultaneous equations in two variables (i), (ii) are solvable.
(1) If (a1 b2 - a2 b1) ≠ 0 for any values of (b1 c2 - b2 c1) and (a2 c1 - a1 c2), we get unique solutions for x and y from equation (iii)
• For examples:
7x + y + 3 = 0 ------------ (i)
2x + 5y – 11 = 0 ------------ (ii)

• Here, a1 = 7, a2 = 2, b1 = 1, b2 = 5, c1 = 3, c2 = -11
and (a1 b2 - a2 b1) = 33 ≠ 0 from equation (iii)
• we get, x = -26/33 , y = 83/33
Therefore, (a1 b2 - a2 b1) ≠ 0, then the simultaneous equations (i), (ii) are always consistent.
(2) If (a1 b2 - a2 b1) = 0 and one of (b1 c2 - b2 c1) and (a2 c1 - a1 c2) is zero (in that case, the other one is also zero), we get,
• a1/a2 = b1/b2 = c1/c2 = k (Let) where k ≠ 0
• that is, a1 = ka2, b1 = kb2 and c1 = kc2 and changed forms of the simultaneous equations are
• ka2x + kb2y + kc2 = 0
a2x + b2y + c2 = 0
But they are two different forms of the same equation; expressing x in terms of y, we get
• x = - b2y + c2/a2

Which indicates that for each definite value of y, there is a definite value of x, in other words, there are infinite number of solutions of the simultaneous equations in this case?
For examples:
7x + y + 3 = 0
14x + 2y + 6 = 0

Here, a1/a2 = b1/b2 = c1/c2 = 1/2
Actually, we get the second equation when the first equation is multiplied by 2.
In fact, there is only one equation and expressing x in term of y, we get:
x = -(y + 3)/7
Some of the solutions in particular:

(3) If (a1 b2 - a2 b1) = 0 and one of (b1 c2 - b2 c1) and (a2 c1 - a1 c2) is non-zero (then the other one is also non-zero) we get,
(let) k = a1/a2 = b1/b2 ≠ c1/c2

That is, a1 = ka2 and b1 = kb2
In this case, the changed forms of simultaneous equations (i) and (ii) are

• ka2x + kb2y + c1 = 0 ………. (v)
• a2x + b2y + c2 = 0 ………. (vi)
and equation (iii) do not give any value of x and y. So the equations are inconsistent.
• At the time of drawing graphs, we will notice that a linear equation in two variables always represents a straight line and the two equations of the forms (v) and (vi) represent two parallel straight lines.
For that reason, they do not have any common point.
• For examples: 7x + y + 3 = 0
14x + 2y - 1 = 0
Here, a1 = 7, b1 = 1, c1 = 3 and a2 = 14, b2 = 2, c2 = -1
• and a1/a2 = b1/b2 ≠ c1/c2
So, the given simultaneous equations are inconsistent.
• From the above discussion, we can arrive at the following conclusions that the solvability of linear simultaneous equations in two variables
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 will be
• (1) Consistent if a1/a2 ≠ b1/b2: in this case, we will get unique solution
(2) Inconsistent, that is, there will be no solution if
a1/a2 = b1/b2 ≠ c1/c2 where c1 ≠ 0, c2 ≠ 0
(3) Consistent having infinite solution if
a1/a2 = b1/b2 = c1/c2 where c1 ≠ 0, c2 ≠ 0

Word Problems on Simultaneous Linear Equations

Problems of different problems with the help of linear simultaneous equations:
We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations.

• In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols.
• Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem.

Worked-out examples for the word problems on simultaneous linear equations:

1. The sum of two number is 14 and their difference is 2. Find the numbers.
Solution:

Let the two numbers be x and y.
x + y = 14………. (i)
x – y = 2………. (ii)
Adding equation (i) and (ii), we get 2x = 16
or, 2x/2 = 16/2 or, x = 16/2
or, x = 8
Substituting the value x in equation (i), we get
8 + y = 14
or, 8 – 8 + y = 14 – 8
or, y = 14 – 8
or, y = 6
Therefore, x = 8 and y = 6
Hence, the two numbers are 6 and 8.

2. In a two digit number. The units digit is thrice the tens digit.
If 36 is added to the number, the digits interchange their place. Find the number.
Solution:

Let the digit in the units place is x
And the digit in the tens place be y.

• Then x = 3y and the number = 10y + x
The number obtained by reversing the digits is 10x + y.
If 36 is added to the number, digits interchange their places,
Therefore, we have 10y + x + 36 = 10x + y
or, 10y – y + x + 36 = 10x + y – y
or, 9y + x – 10x + 36 = 10x – 10x
or, 9y – 9x + 36 = 0 or, 9x – 9y = 36
or, 9(x – y) = 36
or, 9(x – y)/9 = 36/9
• or, x – y = 4 ………. (i)
• Substituting the value of x = 3y in equation (i), we get
3y – y = 4
or, 2y = 4
or, y = 4/2
or, y = 2
• Substituting the value of y = 2 in equation (i),we get
x – 2 = 4
or, x = 4 + 2
or, x = 6
Therefore, the number becomes 26.

3. If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions.
Solution:

Let the fraction be x/y.
If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have
(x + 2)/(y + 2) = 9/10
or, 10(x + 2) = 9(y + 2)
or, 10x + 20 = 9y + 18
or, 10x – 9y + 20 = 9y – 9y + 18
or, 10x – 9x + 20 – 20 = 18 – 20
or, 10x – 9y = -2………. (i)

• If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have
(x – 3)/(y – 3) = 4/5
or, 5(x – 3) = 4(y – 3)
or, 5x – 15 = 4y – 12
or, 5x – 4y – 15 = 4y – 4y – 12
or, 5x – 4y – 15 + 15 = – 12 + 15
or, 5x – 4y = 3………. (ii)
• So, we have 10x – 9y = –2………. (iii)
• and 5x – 4y = 3………. (iv)
• Multiplying both sided of equation (iv) by 2, we get
10x – 8y = 6………. (v)
• Now, solving equation (iii) and (v) , we get
10x – 9y = -2
10x – 8y = 6
- y = - 8
y = 8
• Substituting the value of y in equation (iv)
5x – 4 × (8) = 3
5x – 32 = 3
5x – 32 + 32 = 3 + 32
5x = 35
x = 35/5
x = 7
• Therefore, fraction becomes 7/8.

4. If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son.
Solution:

• Let father’s age be x years
• Son’s ages = y years
Then 2y + x = 56 …………… (i)
And 2x + y = 82 …………… (ii)
Multiplying equation (i) by 2, (2y + x = 56 …………… × 2)we get
• or, 3y/3 = 30/3
or, y = 30/3
or, y = 10 (solution (ii) and (iii) by subtraction)
Substituting the value of y in equation (i), we get;
2 × 10 + x = 56
or, 20 + x = 56
or, 20 – 20 + x = 56 – 20
or, x = 56 – 20

• x = 36
• y=10

5. Two pens and one eraser cost Rs. 35 and 3 pencil and four erasers cost Rs. 65. Find the cost of pencil and eraser separately.
Solution:

Let the cost of pen = x and the cost of eraser = y
Then 2x + y = 35 ……………(i)
And 3x + 4y = 65 ……………(ii)
Multiplying equation (i) by 4, Subtracting (iii) and (ii), we get;
5x = 75
or, 5x/5 = 75/5
or, x = 75/5
or, x = 15
Substituting the value of x = 15 in equation (i) 2x + y = 35 we get;
or, 2 × 15 + y = 35
or, 30 + y = 35
or, y = 35 – 30
or, y = 5
Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.

Solution Of a Linear Equations

Examples on solving linear equations:
1. Solve the equation 2x - 1 = 14 - x and represent the solution graphically.
Solution:

2x - 1 = 14 - x
⇒ 2x + x = 14 + 1
(Transfer -x from right hand side to the left hand side, then negative x changes to positive x.
Similarly again transfer -1 from left hand side to the right hand side, then negative 1 change to positive 1.
Therefore, we arranged the variables in one side and the numbers in the other side.)
⇒ 3x = 15
⇒ 3x/3 = 15/3 (Divide both sides by 3)
⇒ x = 5
Therefore, x = 5 is the solution of the given equation.
The solution may be represented graphically on the number line by graphing linear equations.

2. Solve the equation 10x = 5x + 1/2 and represent the solution graphically.
Solution:

10x = 5x + 1/2
⇒ 10x – 5x = 1/2
(Transfer 5x from right hand side to the left hand side, then positive 5x changes to negative 5x).
⇒ 5x = 1/2
⇒ 5x/5 = 1/2 ÷ 5 (Divide both sides by 5)
⇒ x = 1/2 × 1/5
⇒ x = 1/10
Therefore, x = 1/10 is the solution of the given equation.
The solution may be represented graphically on the number line. 3. Solve the equation 6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3) and verify your answer
Solution:

6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3)
⇒ 18x + 12 + 35x - 30 - 12x = 30x - 5 + 6x - 18
⇒ 18x + 35x - 12x + 12 - 30 = 30x + 6x - 5 - 18
⇒ 41x - 18 = 36x - 23
⇒ 41x - 36x = - 23 + 18
⇒ 5x = -5
⇒ x = -5/5
⇒ x = -1
Therefore, x = -1 is the solution of the given equation.
Now we will verify both the sides of the equation,
6(3x + 2) + 5(7x - 6) - 12x = 5(6x - 1) + 6(x - 3) are equal to each other;
Verification:
L.H.S. = 6(3x + 2) + 5(7x - 6) - 12x
Plug the value of x = -1 we get;
= 6[3 × (-1) + 2] + 5 [7 × (-1) - 6] - 12 × (-1)
= 6[-3 + 2] + 5[-7 - 6] + 12
= 6 × (-1) + 5 (-13) + 12
= - 6 - 65 + 12
= -71 + 12
= -59
Verification:
R.H.S. = 5(6x - 1) + 6(x - 3)
Plug the value of x = - 1, we get
= 5[6 × (-1) - 1] + 6[(-1) - 3]
= 5(-6 - 1) + 6(-1 -3)
= 5 × (-7) + 6 × (-4)
= - 35 - 24
= - 59
Since, L.H.S. = R.H.S. hence verified.

What is cross multiplication?

• The process of multiplying the numerator on the left hand side with the denominator on the right hand side and multiplying the denominator on left hand side with the numerator on right hand side is called cross multiplication.
• And then equating both the products we get the linear equation.
• On solving it we get the value of variable for which L.H.S. = R.H.S. Then, it is an equation of the form.
• (mx + n)/(ox + p) = q/r where m, n, o, p, q, r are numbers and ox + p ≠ 0
⇒ r(mx + n) = q(ox + p)
• It’s an equation in one variable x but it is not a linear equation as L.H.S. is not a linear polynomial.
• We convert this into linear equation by the method of cross multiplication and further solve it step-by-step.

Examples on cross multiplication while solving linear equations:

1. (3x + 4)/5 = (2x - 3)/3
Solution:

(3x + 4)/5 = (2x - 3)/3
On cross multiplication, we get;
⇒ 3(3x + 4) = 5(2x - 3)
⇒ 9x + 12 = 10x - 15
⇒ 9x - 10x = -15 - 12
⇒ -x = -27
⇒ x = 27
Verification:
L.H.S. = (3x + 4)/5
Plug x = 27, we get;
(3 × 27 + 4)/5
= 81 + 4/5
= 85/5
= 17
Verification:
R.H.S. = (2x - 3)/3
Plug x = 27, we get;
(2 × 27 - 3)/3
= 54 - 3/3
= 51/3
= 17
Since, L.H.S. = R.H.S. hence verified.

2. Solve 0.8 - 0.28x = 1.16 - 0.6x
Solution:

0.8 - 0.28x = 1.16 - 0.6x
⇒ 0.6x - 0.28x = 1.16 - 0.8
⇒ 0.32x = 0.36
⇒ x = 0.36/0.32
⇒ x = 36/32
⇒ x = 9/8
Therefore, 9/8 is the required solution.
Verification:
L.H.S. = 0.8 - 0.28x
Plug x = 9/8, we get;
= 0.8 - 0.28 × 9/8
= 8/10 - 28/100 × 9/8
= 8/10 - 63/200
= (160 - 63)/200
= 97/200
Verification:
R.H.S. = 1.16 - 0.6x
= 1.16 - 0.6 × 9/8
= 116/100 - 6/10 × 9/8
= 116/100 - 27/40
= (232 - 135)/200
= 97/200
Since, L.H.S. = R.H.S. hence verified.

Simultaneous Linear Equations

• Two linear equations in two variables taken together are called simultaneous linear equations. The solution of system of simultaneous linear equation is the ordered pair (x, y) which satisfies both the linear equations.

Necessary steps for forming and solving simultaneous linear equations

Let us take a mathematical problem to indicate the necessary steps for forming simultaneous equations:
Example:
In a stationery shop, cost of 3 pencil cutters exceeds the price of 2 pens by Rs2.
Also, total price of 7 pencil cutters and 3 pens is Rs 43.
Follow the steps of instruction along with the method of solution.
Step I:
Indentify the unknown variables; assume one of them as x and the other as y
Here two unknown quantities (variables) are:
Price of each pencil cutter = Rsx
Price of each pen = Rsy
Step II:
Identify the relation between the unknown quantities.
Price of 3 pencil cutter =Rs3x
Price of 2 pens = Rs2y
Therefore, first condition gives: 3x – 2y = 2
Step III:
Express the conditions of the problem in terms of x and y
Again price of 7 pencil cutters = Rs7x
Price of 3 pens = Rs3y
Therefore, second condition gives: 7x + 3y = 43
Simultaneous equations formed from the problems:
3x – 2y = 2 ----------- (i)
7x + 3y = 43 ----------- (ii)

Worked-out problems on solving simultaneous linear equations

1. x + y = 7 ………… (i)
3x - 2y = 11 ………… (ii)
Solution:

The given equations are:
x + y = 7 ………… (i)
3x – 2y = 11 ………… (ii)
From (i) we get y = 7 – x
Now, substituting the value of y in equation (ii), we get;
3x – 2 (7 – x) = 11
or, 3x – 14 + 2x = 11
or, 3x + 2x – 14 = 11
or, 5x – 14 = 11
or, 5x -14 + 14 = 11 + 14 [add 14 in both the sides]
or, 5x = 11 + 14
or, 5x = 25
or, 5x/5 = 25/5 [divide by 5 in both the sides]
or, x = 5
Substituting the value of x in equation (i), we get;
x + y = 7
Put the value of x = 5
or, 5 + y = 7
or, 5 – 5 + y = 7 – 5
or, y = 7 – 5
or, y = 2
Therefore, (5, 2) is the solution of the system of equation x + y = 7 and 3x – 2y = 11

2. Solve the system of equation 2x – 3y = 1 and 3x – 4y = 1.
Solution:

The given equations are:
2x – 3y = 1 ………… (i)
3x – 4y = 1 ………… (ii)
From equation (i), we get;
2x = 1 + 3y
or, x = 1/2(1 + 3y)
Substituting the value of x in equation (ii), we get;
or, 3 × 1/2(1 + 3y) – 4y = 1
or, 3/2 + 9/2y - 4y = 1
or, (9y – 8y)/2 = 1 – 3/2
or, 1/2y = (2 – 3)/2
or, 1/2y = -1/2
or, y = -1/2 × 2/1
or, y = -1
Substituting the value of y in equation (i)
2x – 3 × (-1) = 1
or, 2x + 3 = 1
or, 2x = 1 - 3or, 2x = -2
or, x = -2/2
or, x = -1
Therefore, x = -1 and y = -1 is the solution of the system of equation
2x – 3y = 1 and 3x – 4y = 1.

3.Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 4.37 (ii) x – 4 = 3 y (iii) 4 = 5x – 3y (iv) 2x = y
Solution :

(i) 2x + 3y = 4.37 can be written as 2x + 3y – 4.37 = 0. Here a = 2, b = 3 and c = – 4.37.
(ii) The equation x – 4 = 3 y can be written as x – 3 y – 4 = 0. Here a = 1, b = – 3 and c = – 4.
(iii) The equation 4 = 5x – 3y can be written as 5x – 3y – 4 = 0. Here a = 5, b = –3 and c = – 4.
Do you agree that it can also be written as –5x + 3y + 4 = 0 ?
In this case a = –5, b = 3 and c = 4.
(iv) The equation 2x = y can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and c = 0.
Equations of the type ax + b = 0 are also examples of linear equations in
two variables because they can be expressed as ax + 0.y + b = 0
For example, 4 – 3x = 0 can be written as –3x + 0.y + 4 = 0

4 .Write each of the following as an equation in two variables:
(i) x = –5
(ii) y = 2
(iii) 2x = 3
(iv) 5y = 2
Solution :

(i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0.
(ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0.
(iii) 2x = 3 can be written as 2x + 0.y – 3 = 0.
(iv) 5y = 2 can be written as 0.x + 5y – 2 = 0.

• Let us consider the equation 2x + 3y = 12. Here, x = 3 and y = 2 is a solution because when you substitute x = 3 and y = 2 in the equation above, you find that
2x + 3y = (2 × 3) + (3 × 2) = 12
• This solution is written as an ordered pair (3, 2), first writing the value for x and then the value for y. Similarly, (0, 4) is also a solution for the equation above.
• On the other hand, (1, 4) is not a solution of 2x + 3y = 12, because on putting x = 1 and y = 4 we get 2x + 3y = 14, which is not 12. Note that (0, 4) is a solution but not (4, 0).
• You have seen at least two solutions for 2x + 3y = 12, i.e., (3, 2),(0, 4)and (6, 0) .
• In fact, we can get many many solutions in the following way. Pick a value of your choice for x (say x = 2) in 2x + 3y = 12. Then the equation reduces to 4 + 3y = 12,
• which is a linear equation in one variable. On solving this, you get y = 8/3 . So (2, 8 /3) is another solution of 2x + 3y = 12.
Similarly, choosing x = – 5, you find that the equation becomes –10 + 3y = 12.
• This gives y = 22/ 3 . So, [-5,22/3] is another solution of 2x + 3y = 12. So there is no end to different solutions of a linear equation in two variables.
• That is, a linear equation in two variables has infinitely many solution

Coordinate Graph

In graphing a coordinate graph is a grid with x axes and y axes crossing through the center and the intersecting point is marked by a dot.
Coordinate Axes:

• On a graph-paper the two mutually drawn perpendicular straight lines X’OX and YOY’, intersecting each other at the point O.
These perpendicular straight lines are known ascoordinate axes.
• The line X’OX is called the x-axis and the line YOY’ is called the y-axis, while the point 0 is called the origin.
• The plane of the graph paper containing both the coordinate axes (x-axes and y-axes) is called the Cartesian plane.
• On right-hand side of the y-axis, every end-point of a square on the x-axis, represents a positive integer, as shown in the figure.
• On left-hand side of the y-axis, every end-point of a square on the x-axis, represents a negative integer, as shown in the figure.
• Above the x-axis, every end-point of a square on the y-axis, represents a positive integer, as shown in the figure.
• Below the x-axis, every end-point of a square on the y-axis,, represents a negative integer, as shown in the figure.

Coordinates of a Point

On a graph paper draw X'OX and YOY'. Here X'OX and YOY’ be the two coordinate axes. • Then mark a point on the graph and name the point as P such that P is at the perpendicular distance of a units from the y-axis and again similarly b units from the x-axis. Then, we denote that the coordinates of P are P(a, b)
• From the above discussion, a is called theabscissa or x-coordinate of P and b is called the ordinate or y-coordinate of P.

Examples to get coordinates of a point

1. In the adjoining figure, to find the co-ordinate of point P the distance of P from y-axis is 5 units and the distance of P from x axis is 5 units. Therefore the co-ordinates of point P are (5, 5).

2. In the adjoining figure, to find the co-ordinate of point P the distance of P from y-axis is 7 units and the distance of P from x axis is 4 units. Therefore the co-ordinates of point P are (7, 4).

• Learn all four quadrants of a coordinate system.
• The plane of the graph paper is divided into four regions by the coordinate axes and the four regions are called quadrants.

What are the Quadrants in the Coordinate Plane? The region XOY is called the I quadrant.
In first quadrant, both x and y co-ordinate are positive.
• The region X’OY’ is called the II quadrant.
In second quadrant x co-ordinate is negative and y co-ordinate is positive.
• The region X’OY’ is called the III quadrant.
In third quadrant both x and y co-ordinate are negative.
• The region XOY’ is called the IV quadrant.
In fourth quadrant x co-ordinate is positive and y co-ordinate is negative.
According to the quadrants the signs of the co-ordinate of a point are of the form:
(+, +) in the first quadrant
(-, +) in the second quadrant
(-, -) in the third quadrant
(+, -) in the fourth quadrant
The table will help us to remember the region of the quadrants and theirs signs of coordinates.
region quadrant signs of co ordinates
XOY I (+, +)
YOX’ II (-, +)
X’OY’ III (-, -)
Y’OX IV (+, -)

Signs of Coordinates

Here we will learn about the signs of coordinates.
→X′OX and →Y′OY represent the co-ordinate axes.
.The ray →OX is taken as positive x-axis. So, any distance along →OX will be taken as positive and the ray →OX′ is taken as negative x-axis. So, any distance move along →OX′ will be taken as negative.
Similarly, ray →OY taken as positive y-axis.
So, the distance moved along →OY will be taken as positive and →OY′ is taken as negative y-axis.
So, the distance moved along →OY′ will be taken as negative.  The graph will help us to understand the convention of the signs of coordinates. (i) The co-ordinates of any point lies in the first quadrant have both the abscissa and ordinate are positive i.e. (+, +).
(ii) The co-ordinates of any point lies in the second quadrant have the abscissa negative and ordinate positive i.e. (-, +).
(iii) The co-ordinates of any point lies in the third quadrant have both the abscissa and ordinate are negative i.e. (-, -).
(iv) The co-ordinates of any point lies in the fourth quadrant have the abscissa positive and ordinate negative i.e. (+, -).

How to determine the coordinates of a point in a plane?
Length of bar(ON) is called the x- co-ordinate or abscissa of point P. Here bar(ON) = 2 units.
Length of bar(OM) is called the y-co-ordinate or ordinate of P. Here bar(OM) = 4 units. Thus, the co-ordinates of point P are (2, 4) which is called an ordered pair.
So, the positions of the coordinates of a point in a plane cannot be interchanged as (4, 2).
Remember, if the distance of P from y-axis is ‘a’ and units the distance of P from the x-axis is ‘b’ units then the co-ordinates of point P are (a, b) where a denotes the x-co-ordinate orabscissa and b denotes the y-co-ordinate or ordinate.
Thus, we can define abscissa as distance of P from y-axis and ordinate as the distance of P from x-axis.
Point on x-axis: If we take any point on x-axis, then the distance of this point from x-axis is zero i.e.,
y-co-ordinate of every point on x-axis is zero.
Therefore, the co-ordinates of a point on x-axis are of the form (x, 0)
Point on y-axis: If we take any point on y-axis, then the distance of this point from y-axis is zero i.e.,
x-co-ordinate of every point on y-axis is zero.
Therefore, the co-ordinates of a point on y-axis are of the form (0, y).

Plot Points on Coordinate Graph

How do you plot points on coordinate graph?

• We draw co-ordinate axes X'OX, extended ‘side-by side’ and Y'OY, extended ‘above and below’ and take x-axis and y-axis respectively.
• Then we have to decide how any sides of the small square may be taken as one unit such that one centimeter represents one unit on both the axes. According to that unit and sign of abscissa and ordinate, the location of the point is found

Example to plot points on a coordinate graph paper

1. Plot point P(2, 4) on the graph.
Solution:

• On the graph paper X'OX and Y'OY are indicated as x-axis and y-axis respectively. Length of one sides of the square (i.e. 1 div) is taken as unit.
• In point P(2, 4) we observe that both the co-ordinates are positive so they will lie in the first quadrant.
• Count 2 units along x-axis to the right of origin. Draw a line BA ┴ XOX'.
• Now, count 4 units along y-axis upward. Draw line CD ┴ YOY'.
• Both these lines intersect at point P.
• Therefore, the co-ordinates of point P are (2, 4).

2. Plot point M(-5, -3) on the graph.
Solution:

• On the graph paper X'OX and Y'OY are indicated as x-axis and y-axis respectively.
• Length of one sides of the square (i.e. 1 div) is taken as unit.
• In point M(-5, -3) we observe that both the co-ordinates i.e. abscissa (-5) and ordinate (-3) are negative so they will lie in the third quadrant.
• Count 5 units along x-axis to the left of origin. Draw a line QP ┴ X'OX.
• Now, count 3 units along y-axis downward. Draw line RS ┴ Y'OY.
• Both these lines intersect at point M.
• Therefore, the co-ordinates of point M are (-5, -3).

Note:
Any point on x-axis: The co-ordinates of any point on the x-axis are of the form (x, 0).
Therefore the y co-ordinate of every point on x axis is zero.
For example: (2, 0), (7, 0), (5, 0), (-2, 0), (-7, 0), (-2, 0) are the points which lie on x –axis.
Any point on y-axis: The co-ordinates of any point on the y-axis are of the form (0, y).
Therefore the x co-ordinate of every point on y axis is zero.
For example: (0, 1), (0, 4), (0, 6), (0, -1), (0, -4), (0, -6) are the points which lie on y-axis.

Graph of Linear Equation

Linear equation in two variables is represented graphically by the line whose points give the collection of solutions of the equation.
This is called graphing linear equation.
Properties for graphing linear equation:
1. Linear equations have infinitely many solutions.
2. Every point (h, k) on the line AB gives the solution x = h and y = k.
3. Every point which lies on AB satisfies the equation of AB.
4. To draw an exact line on the graph paper you can plot as many points you like, but it is necessary to plot minimum three points.

Method to draw the graph of linear equation in two variables

1. Convert the given equation in the form of y = mx + b (slope intercept form).
2. Apply trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation.
3. Plot these points on the graph paper.
4. Join the points marked on the graph paper to get a straight line which represent the given equation graphically.

Note:
1. Linear equation in two variables has infinitely many solutions.
2. A graph of linear equation is always a straight line.
3. Every point on the straight line is the solution of the linear equation.
4. Equation of y-axis is x = 0. The standard form of this equation is x + 0.y = 0.
5. Equation of x-axis is y = 0. The standard form of this equation is 0.x + y = 0.
6. x = a is a graph of straight line parallel to y-axis and standard form of this equation is x + 0.y = a
7. y = b is a graph of straight line parallel to x-axis and standard form of this equation is 0.x + y = b.
8. The equation y = mx is always passing through the origin (0, 0).

Learn the steps for graphing linear equation in two variables:

1. Draw the graph of the linear equation y = 2x.
Solution:

The given linear equation y = 2x is already in the form of y = mx + b [here b = 0].
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 2x.
When the value of x = 0, then y = 2 × 0 = 0
When the value of x = 1, then y = 2 × 1 = 2
When the value of x = 3, then y = 2 × 3 = 6
When the value of x = -1, then y = 2 × -1 = -2
When the value of x = -2, then y = 2 × -2 = -4

Arrange the values of the linear equation y = 2x in the table.

 x y 1 2 3 4 4 3 2 1

Now, plot the points P (0, 0), Q (1, 2), R (2, 4), S (3, 6), T (-1, -2), U (-2, -4) on the graph paper. Join the points of P, Q, R, S, T and U.
We get a straight line passing through origin.
This straight line is the graph of the equation y = 2x.

2. Draw the graph of the equation 4x - y = 3.
Solution:

The given linear equation 4x - y = 3.
Now convert the given equation in the form of y = mx + b
4x - y = 3
⇒ 4x - 4x - y = - 4x + 3
⇒ - y = - 4x + 3
⇒ y = 4x - 3
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 4x - 3.
When the value of x = 0, then y = (4 × 0) - 3 = - 3
When the value of x = 1, then y = (4 × 1) - 3 = 1
When the value of x = 2, then y = (4 × 2) - 3 = 5
Arrange these value of the linear equation y = 4x - 3 in the table.

 x y 0 1 2 -3 1 5

Now, plot the point P (0, -3), Q (1, 1), R(2, 5) on the graph. Join the points of P, Q and R.
We get a straight line passing through origin.
This straight line is the graph of linear equation 4x - y = 3.

Simultaneous Equations Graphically

How to solve a pair of simultaneous equations graphically?
In order to solve a pair of simultaneous equations graphically, we first draw the graph of the two equations simultaneously.
We get two straight lines intersecting each other at a common point.
This common point intersection of two lines gives the solution of the pair of simultaneous equations.
How to find the solution of simultaneous linear equations by graphing method?
Suppose the given system of linear equations is
p1x + q1y + r1 = 0 ------------- Equation (1)
p2x + q2y + r2 = 0 ------------- Equation (2)

Now draw the both given linear equations on a graph paper.
Graph of linear equation is always a straight line.
Let the line L1 represent the graph of p1x + q1y + r1 = 0 and L2 represent the graph of p2x + q2y + r2 = 0.
Then by solving the simultaneous equations graphically we can get three possible solutions.

1. When the two lines L1 and L2 intersect at a single point:
Then the two lines cross at a single point P (h, k), as shown in the graph. Therefore, x = h and y = k is the unique solution of the given two equations.
This system is called independent.

Solved examples of a pairs of simultaneous equations graphically where the two lines intersect at a single point

Solve graphically the system if linear equation x + y = 5 and x – y = 3.
Solution

On a graph paper to solve simultaneous equations graphically, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis respectively.
x + y = 5 ------------ Equation (1)
Now convert the given equation in the form of y = mx + b
x + y = 5
⇒ x - x + y = - x + 5
⇒ y = - x + 5
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = -x + 5.
When the value of x = 1 then y = -1 + 5 = 4
When the value of x = 2 then y = -2 + 5 = 3
When the value of x = 3 then y = -3 + 5 = 2
When the value of x = 4 then y = -4 + 5 = 1
Arrange these value of the linear equation y = -x + 5 in the table 1.

 x y 1 2 3 4 4 3 2 1
Table 1

Now plot the points of the equation x + y = 5; A (1, 4), B (2, 3), C (3, 2), D (4, 1) on the graph paper.
Join the points of A, B, C and D; to get the graph line AD.
Thus line AD is the graph of x + y = 5.
x – y = 3 ------------ Equation (2)
Now convert the given equation in the form of y = mx + b
x – y = 3
⇒ x – x – y = – x + 3
⇒ -y = -x + 3
⇒ y = x – 3
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = x - 3.
When the value of x = 1 then y = 1 – 3 = -2
When the value of x = 2 then y = 2 – 3 = -1
When the value of x = 3 then y = 3 – 3 = 0
When the value of x = 4 then y = 4 – 3 = 1
Arrange these value of the linear equation y = x - 3 in the table 2

 x y 1 2 3 4 -2 -1 0 1

Table 2

Now plot the points of the equation x – y = 3; P (1, -2), Q (2, -1), R (3, 0), S (4, 1) on the graph paper.
Join the points of P, Q, R and S; to get the graph line PS.
Thus line PS is the graph of x – y = 3. We get two straight lines intersecting each other at (4, 1).
Therefore, x = 4 and y = 1 is the solution of the given system of equation.

2. When the two lines L1 and L2 are coincident:
Then the two equations actually be the same line, as shown in the graph. Therefore, the given two equations have infinitely many solutions.
This system is called dependent.

Solved examples of a pairs of simultaneous equations graphically where the two lines coincidentally be the same line

Solve graphically the system if linear equation 3x - y = 2 and 9x – 3y = 6.
Solution:

• On a graph paper to solve simultaneous equations graphically, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis respectively.
• 3x - y = 2 ------------ Equation (1)
• Now convert the given equation in the form of y = mx + b
• y = 3x - 2
• Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 3x - 2.
• When the value of x = -1 then y = 3(-1) - 2 = -5
• When the value of x = 0 then y = 3(0) - 2 = -2
When the value of x = 2 then y = 3(2) - 2 = 4
Arrange these value of the linear equation y = 3x - 2 in the table 1.
 x y -1 0 2 -5 -2 4

Table 1

Now plot the points of the equation 3x - y = 2; A (-1, -5), B (0, -2), C (2, 4) on the graph paper.
Join the points of A, B and C; to get the graph line AC.
Thus line AC is the graph of 3x - y = 2.
9x – 3y = 6 ------------ Equation (2)
Now convert the given equation in the form of y = mx + b
y = (9x – 6)/3
⇒ y = 9x/3 – 6/3
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 9x/3 – 6/3.
When the value of x = -2 then y = -8
When the value of x = 1 then y = 1
When the value of x = 2 then y = 4
Arrange these value of the linear equation y = 9x/3 – 6/3 in the table 2.

 x y -2 1 2 -8 1 4

Table 2
Now plot the points of the equation 9x – 3y = 6; P (1, -2), Q (2, -1), R (3, 0) on the graph paper.
Join the points of P, Q and R; to get the graph line PR.
Thus line PR is the graph of 9x – 3y = 6. We find that the two straight lines AC and PR coincides each other.
Therefore, the given system of equations has an infinite number of solutions.

3. When the two lines L1 and L2 are parallel:
Then the two equations have no common solutions, as shown in the graph. This system is called inconsistent.

Solved examples of a pairs of simultaneous equations graphically when the two lines are parallel

Solve graphically the system if linear equation 2x - 3y = 5 and 6y – 4x = 3.
Solution:

On a graph paper to solve simultaneous equations graphically, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis respectively.
2x - 3y = 5 ------------ Equation (1)
Now convert the given equation in the form of y = mx + b
2x - 3y = 5
⇒ 3y = (2x – 5)
⇒ y = (2x – 5)/3
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = (2x – 5)/3
When the value of x = -2 then y = -3
When the value of x = 1 then y = -1
When the value of x = 4 then y = 1
Arrange these value of the linear equation y = (2x – 5)/3 in the table 1.

 x y -2 1 4 -3 -1 1

Table1

Now plot the points of the equation 2x - 3y = 5; A (-2, -3), B (1, -1) and C (4, 1) on the graph paper.
Join the points of A, B and C; to get the graph line AC.
Thus line AC is the graph of 2x - 3y = 5.
6y – 4x = 3 ------------ Equation (2)
Now convert the given equation in the form of y = mx + b
6y – 4x = 3
⇒ 6y = (3 + 4x)
⇒ y = (3 + 4x)/6
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = (3 + 4x)/6.
When the value of x = -3 then y = -3/2
When the value of x = 0 then y = 1/2
When the value of x = 3 then y = 5/2
Arrange these value of the linear equation y = (3 + 4x)/6 in the table 2.

 x y -3 0 3 -3/2 1/2 5/2

Table 2

Now plot the points of the equation 6y – 4x = 3; P (-3, -3/2), Q (0, 1/2), R (3, 5/2) on the graph paper.
Join the points of P, Q and R; to get the graph line PR.
Thus line PR is the graph of 6y – 4x = 3 We get from the graph that two straight lines are parallel to each other.
Therefore, the given system of equations has no solutions.
Thus we conclude that to solve a pair of simultaneous equations graphically then the possible outcomes can be:
(i) a unique solution if the graph lines intersect at a point.
(ii) infinitely many solutions if the two graph lines coincides.
(iii) no solution if the two graph lines are parallel.

Graphs of Simple Function

Here we will learn about the graphs of simple functions that is multiples of different numbers

The example will help us to understand the graph of multiples of different numbers.
1. Draw the graph of the function y = 3x.
2. From the graph, find the value of y, when
(a) x = 4
(b) x = 5
Solution:

1. The given function is y = 3x.
For some different values of x, the corresponding values of y are given below.

 x y = 3x 0 1 2 3 0 3 6 9

On a graph paper plot the points O (0, 0), A (1, 3), B (2, 6) and C (3, 9) and join them successively to obtain the required graph. 2. Reading off from the graph of simple function:
(a) On the x-axis, take the point L at x = 4.
Draw LP ⊥ x-axis, meeting the graph at P.
Clearly, PL = 12 units.
Therefore, x = 4 ⇒ y = 12.
(b) On the x-axis, take the point M at x = 5.
Draw MQ ⊥ x-axis, meeting the graph at Q.
Clearly MQ = 15 units
Therefore, x = 5 ⇒ y = 15

Graph of Perimeter vs. Length of the Side of a Square
Here we will learn about the graph of perimeter vs. length of the side of a square
For a square we have, perimeter = (4 × side of the square).

1. Consider the relation between the perimeter and the side of a square, given by P = 4a.
Draw a graph of the above function.
2. From the graph, find the value of P. when
(a) a = 5
(b) a = 6.
Solution:

1. The given function is P = 4a.
For different values of a, the corresponding values of P are given below.

 a p = 4a 0 1 2 3 4 0 4 8 12 16

On a graph paper, plot the points O (0, 0), A (1, 4), B (2, 8), C (3, 12) and D (4, 16).
Join these points to get the required graph line, shown below. 2. Reading off from the graph of perimeter vs. length of the side of a square:
(a) On the x-axis, take the point L at a = 5.
Draw LE ⊥ x-axis, meeting the graph at E.
Clearly, EL = 20 units.
Therefore, a = 5 ⇒ P = 20.
Thus, when a = 5, then P = 20.
(b) On the x-axis, take the point M at a = 6.
Draw MF ⊥ x-axis, meeting the graph at F.
Clearly, MF = 24 units.
Therefore, a = 6 ⇒ P = 24.
Thus, when a = 6, then P = 24.

Graph of Area vs. Side of a Square

Here we will learn about the graph of area vs. side of a square.
We know that area of a square = (side)2. Thus, A = x2.

(a) Consider the relation between the area and the side of a square, given by A = x2.
Draw a graph of the above function.
(b) From the graph, find the value of A, when x = 4.
Solution:

The given function is A = x2.
For different values of x, we get the corresponding value of A.
x= 0 ⇒ A = 02 = 0
x= 1 ⇒ A = 12 = 1
x = 2 ⇒ A = 22 = 4
x = 3 ⇒ A = 32 = 9

 x A = X2 0 1 2 3 0 1 4 9

Thus, we have the points O (0, 0), B (1, 1), C (2, 4) and D (3, 9).
Plot these points on a graph paper and join them successively to obtain the required graph given below. 2. Reading off from the graph of area vs. side of a square:
On the x-axis, take the point L at x = 4.
Draw LP ⊥ x-axis, meeting the given graph at P.
Clearly, PL = 16 units.
Therefore, x = 4 ⇒ A = 16.
Thus, when x = 4 units, then A = 16 sq. units

Graph of Simple Interest vs. Number of Years

Here we will learn about the graph of simple interest vs. number of years.
Simple interest on a certain sum is \$ 40 per year.
Then, S = 40 × x, where x is the number of years.

1. Draw a graph of the above function.
2. From the graph find the value of 5, when
(a) x = 5 (b) x = 6
Solution:

1. The given function is S = 40 × x.
Putting x = 1, 2, 3, 4 successively and getting the corresponding value of S.
we get the table given below.

 x s =40 x X 1 2 3 4 40 80 120 160

Along the x-axis:
Take 1 small square = 1 unit.
Along the y-axis:
Take 1 small square = 10 units.
Now, on a graph paper, plot the points

A (1, 40), B (2, 80), C (3, 120) and D (4, 160).
Join them successively to get the required graph line, shown below. 2. Reading off from the graph of simple interest vs. number of years:
(a) On the x-axis, take the point L at x = 5.
Draw LP ⊥ x-axis, meeting the graph at P.
Clearly, PL, = 200 units.
Therefore, x = 5 ⇒ S = 200. ,
Thus, when x = 5, then S = 200.
(b) On the x-axis, take the point M at x = 6.
Draw QM ⊥ x-axis, meeting the graph at Q.
Clearly, QM = 240 units.
Therefore, x = 6 ⇒ S = 240.
Thus, when x = 6, then S = 240.

Graph of Distance vs. Time

To make a graph the information given below is the distance vs. time.
d = 5t. Here, d = distance and t = time.
(i) Find the distance covered when time is 3 sec.
(ii) Find the distance covered in 5 seconds.
(iii) Find the time in which the body covered 30 m.
Solution:
We know d = 5t
When the value of t = 1 then d = 5(1) = 5
When the value of t = 2 then d = 5(2) = 10
When the value of t = 3 then d = 5(3) = 15
When the value of t = 4 then d = 5(4) = 20
When the value of t = 5 then d = 5(5) = 25
When the value of t = 6 then d = 5(6) = 30
When the value of t = 7 then d = 5(7) = 35
Arrange the values of d and t in the table.

 t d 1 2 3 4 5 6 7 5 10 15 20 25 30 35
Now plot the points (1, 5), (2, 10), (3, 15), (4, 20), (5, 25), (6, 30), (7, 35) of the equation d = 5t on the graph paper. i.From the graph we observe that in 3 seconds the distance covered is 15 meters.
(ii) From the graph we observe that in 5 seconds the distance covered is 25 meters.
(iii) From the graph we observe that the body covered 30 meters in 6 seconds.

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