Algebra Chapter 11 Graph
What is a graph
To construct a graph, some sort of paper is used. In the paper, there are some horizontal and vertical equidistant parallel straight lines. As a result of this, the paper is divided into some equal squares.
This is called graph paper. The side of the smallest square is either 1/10th of an inch or 1 mm. The distance is measured by taking as a unit of length the side of one or more than one square.
Usually, there is a bold line after each 10 smallest squares (i.e., after each 1 inch or 1 cm), x-axis and y-axis are drawn along a bold horizontal line and a bold vertical line respectively with the help of a pencil and a scale.
After this, different points are plotted. The straight line or curve obtained by joining these points is a graph. Of course, in our present discussion, we shall consider only a straight line but no curve.
Plotting of points in the graph paper
Let, we have to plot the four points A (7, 8), B (- 6, 9), C (-5, – 6), and D(10, -9) on the graph paper. We draw the horizontal line XOX’ as-the x-axis. We draw the vertical line YOY’ as the y-axis.
The point of intersection of these two lines, that is, point 0, is called the origin. Each axis has a scale. Let us take 1 unit length (1 small square box) = 1 along both axes.
Numbers to the right of the origin along the x-axis are positive, while numbers to the left of the origin are negative. Similarly, numbers above the origin along the y-axis are positive, while numbers below the origin are negative.
1. At first moving the 7 smallest divisions towards the right along OX from the origin 0(0,0) and then moving the 8 smallest divisions upwards parallel to OY point A (7, 8) will be obtained. 7 is called the abscissa (denoted by x) or the x-coordinate of point A. 8 is called the ordinate (denoted by y) or the y-coordinate of the point A. (7,8) are the coordinates of point A, represented as A(7,8).
2. At first moving the 6 smallest divisions towards left along OX’ from the origin 0(0,0) and then moving the 9 smallest divisions upwards parallel to OY point B(- 6,9) will be obtained.
3. At first moving the 5 smallest divisions towards left along OX’ from the origin 0(0,0) and then moving the 6 smallest divisions downwards parallel to OY’ point C(-5, – 6) will be obtained.
4. At first moving the 10 smallest divisions towards the right along OX from the origin 0(0,0) and then moving the 9 smallest divisions downwards parallel to OY’ the point D (10,-9) will be obtained.
Read And Learn More WBBSE Solutions For Class 8 Maths
Some important information
1. Co-ordinates of the origin are (0,0).
2. The y coordinates of all points on the x-axis are zeroes.
3. The x coordinates of all points on the y-axis are zeroes.
4. The equation of the x-axis is y = 0.
5. The equation of the y-axis is x = 0.
6. The coordinate of each point on any straight line parallel to the x-axis is equal. Hence, the equation of any straight line parallel to the x-axis is y = c, where c is a constant.
7. The x-coordinate of each point on any straight line parallel to the y-axis is equal. Hence, the equation of any straight line parallel to the y-axis is x = c, where c is a constant.
8. The x-axis and the y-axis divide the coordinate plane into four regions, called quadrants.
1. In the first quadrant (or XOY), both the x-coordinate and the y-coordinate of a point are positive.
1st quadrant: x > 0, y > 0
2. In the second quadrant (or YOX’), the x-coordinate of a point is negative and the y-coordinate of the point is positive.
2nd quadrant: x < 0, y > 0
3. In the third quadrant (or XOY), both the x-coordinate and the y- coordinate of a point are negative.
3rd quadrant: x < 0, y < 0
4. In the fourth quadrant (or YOX), the x-coordinate of a point is positive, while the y-coordinate of the point is negative.
4th quadrant: x > 0, y < 0.
WBBSE Class 8 Graphs Notes
Algebra Chapter 11 Graph Drawing of graphs of some simple linear equations
Example 1
Draw the graphs of x = 5 and y = -7 in a figure. Find the point of intersection of those straight lines.
Solution :
Given x = 5 And y = -7
From the first equation, we get x = 5. That means, for any value of y the value of x will be 5. Therefore, by satisfying this equation we get the following values
| X | 5 | 5 | 5 |
| y | 1 | -6 | 8 |
From the second equation we get, y = -7.
That means, for any value of x the value will be -7. Therefore, by satisfying this equation we get the following values :
| X | 8 | -1 | -8 |
| y | -7 | -7 | -7 |
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the x-axis and y-axis respectively. Taking one side of the smallest square as the unit of length the points (5,1), (5, – 6), and (5,8) are plotted on the graph paper. Joining these points with a scale and producing both ways the graph AB of equation
x = 5 is obtained.
Again, with the same pair of axes the points (8, -7), (-1, -7), and (-8, -7) are plotted on the graph paper. Joining these points with a scale and producing both ways the graph CD of the equation
y = -7 is obtained.
AB and CD intersect each other at the point (5, -7). Hence, the required point of intersection is (5, -7).
Types of Graphs for Class 8
Example 2
Draw the graph of the straight line y = 3x + 2.
Solution :
The given equation is, y = 3x + 2. From this equation we get,
| X | 0 | 2 | -2 |
| y | 2 | 8 | -4 |
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the x-axis and y-axis respectively.
Taking one side of the smallest square as the unit of length the points (0, 2), (2, 8), (-2, — 4) are plotted on the graph paper. Joining these points with a scale and producing both ways the graph of the given straight line is obtained.
Example 3
Draw the graph of the equation x/a + y/2 = 1. Find the portion of this graph intercepted between the axes.
Solution:
The given equation is, x/4 + y/5 = 1
or, 5x + 4y / 20 = 1
or, 5x + 4y = 20
or, 5x = 20 – 4y
or, x = 20 – 4y / 5
From the equation we get,
| X | 4 | 0 | 8 |
| y | 0 | 5 | -5 |
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the x-axis and y-axis respectively.
Taking one side of the smallest square as the unit of length the points (4, 0), (0, 5), (8, – 5) are plotted on the graph paper. Joining these points with a scale and producing both ways the^ graph, of the given straight line, is obtained. * > ?iV Let, the straight line intersects the x-axis at A and the y-axis at B.
Here’, OA = 4 units and OB = b units
∴ AB = √4² + 5² units
= √16 + 25 units
= √41 units
The required length is √41 units.
Understanding Coordinate Plane in Algebra
Example 4
Draw the graph of the equation x/5 – y/6 = 1. Find the area of the triangle formed by the graph and the axes.
Solution:
The given equation is,
x/5 – y/6 = 1
or, 6x – 5y / 30 = 1
or, 6x – 5y = 30
or, 6x = 5y + 30
or, x = 5y + 30 / 6
From this equation we get,
| X | 5 | 10 | 0 |
| y | 0 | 6 | -6 |
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the x-axis and y-axis respectively.
Taking one side of the smallest square as the unit of length the points (5,0), (10, 6) and (0, – 6) are plotted on the graph paper. Joining these points with a scale and producing both ways the graph of the given straight line is obtained.
Let, the straight line intersects the x-axis at A and the negative y-axis at B.
Here, OA = 5 units and OB = 6 units.
.’. The required area of the triangle
= 1/2 x OA x OB sq. units.
=1/2 x 5 x 6 sq. units. = 15 sq. units.
The required area of the triangle is 15 sq. units.
Example 5
Find the graph of the function x+5 / 3. From the graph find the value of the function when x = 4. For what value of x, the value of the function will be zero.
Solution:
Drawing of the graph of the function x + 5 / 3 means drawing of y = x + 5 / 3.
Now, from this equation we, get
| X | 1 | 7 | -5 |
| y | 2 | 4 | 0 |
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the x-axis and y-axis respectively.
Taking one side of the smallest square as the unit of length the points (1, 2), (7, 4), and (-5, 0) are plotted on the graph paper. Joining these points with a scale and producing both ways the graph of the given function is obtained.
From the graph, it is seen that, if x = 4, the value of the function is 3 and the value of the function is zero when x = -5.
Practice Problems on Graphing Linear Equations
Example 6
Solve with the help of the graph: 4x + 3y = 24 , 3x – 4y = – 7
Solution :
Given
4x + 3y = 24 , 3x – 4y = – 7.
The first equation is
4x + 3y = 24
or, 4x = 24 – 3y
or, x = 24 – 3y / 4
From this equation we get,
| X | 6 | 3 | 9 |
| y | 0 | 4 | -4 |
Second equation is, 3x – 4y = -7
or, 3x = 4y – 7
or, x = 4y -7 / 3
From this equation we get,
| X | -1 | 3 | -5 |
| y | 1 | 4 | -2 |
Two mutually perpendicular straight lines XOX and YOY’ are taken as the x-axis and y-axis respectively. Taking one side of the smallest square as the unit of length the points (6,0), (3, 4), and (9, – 4) obtained from the first equation are plotted on the graph paper. Joining these points with a scale and producing both ways the graph AB of the first equation is obtained.
Again, with the same pair of axes and the same unit the points (—1, 1), (3, 4), (-5, -2) obtained from the second equation are plotted on the graph paper. Joining these points with a scale and producing both ways the graph CD of the second equation is obtained.
The straight lines AB and CD intersect each other at point P. The coordinates of point P are (3, 4).
The required solution is x = 3, y = 4.
Example 7
Draw the graphs of the equations 4s + 3y = 11 and 5x – y = 9. From this find their point of intersection.
Solution:
Given 4s + 3y = 11 And 5x – y = 9.
The first equation is, 4x + 3y = 11
or, 4x = 11 – 3y
or, x = 11 – 3y / 4
From this equation we get,
| X | 2 | -1 | 5 |
| y | 1 | 5 | -3 |
The second equation is,
5x – y = 9 or, 5x = y + 9
or, x = y + 9 / 5
From this equation we get,
| X | 2 | 3 | 4 |
| y | 1 | 6 | 11 |
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the x-axis and y-axis respectively. Taking one side of the smallest square as the unit of length the points (2, 1), (—1, 5), and (5, —3) obtained from the first equation are plotted on the graph paper. Joining these points with a scale and producing both ways the graph AB of the first equation is obtained. Again, with the same pair of axes and the same unit the points (2, 1), (3, 6), (4, 11) obtained from the second equation are plotted on the graph paper. Joining these points with a scale and producing both ways the graph CD of the second equation is obtained. The straight lines AB and CD intersect each other at point P. The coordinates of point P are (2, 1).
Short Notes on Interpreting Graphs
Example 8
Find graphically the area of the triangle obtained by joining the three points (- 4, 5), (10, 8), and (-10, 11).
Solution:
Two mutually perpendicular straight lines XOX’ and YOY’ are taken as the y-axis and y- axis respectively.
Taking one side of the smallest square as the unit of length the given points are plotted on graph paper. The given points are named P, Q, and R. After this, joining the points P(- 4, 5), Q(10, 8), and R(-10, 11) the triangle PQR is obtained. The area of the triangle PQR is to be determined.
The straight lines LM and RN are drawn parallel to the x-axis through the points P and R respectively. The straight lines NM and RL are drawn parallel to the y-axis through the points Q and R. Then RLMN is a rectangle.
Now, the area of the triangle PQR = area of the rectangle RLMN – an area of the triangle RLP – an area of the triangle QPM – an area of the triangle QNR.
= (20 x 6 – 1/2 x 6 x 6 – 1/2 x 14 x 3 – 1/2 x 20 x 3) sq. units
= (120 -18 – 21 – 30) sq. units.
= 51 sq. units.
Application of graph in practical problem
We have seen that it is possible to solve two simultaneous equations with the help of a graph. It is possible to solve some special types of practical problems by the application of graphs. Of course, the range of such types of problems is limited. Usually, the application of a graph in practical problems is possible when both x and y are positive. From the graph, we may find the value of y for any value of x and also the value of x for any value of y.
Example 1
If the price of 1 kg of rice is ₹ 15, calculate with the help of the graph
1. What is the price of 2 kg of rice?
2. What quantity of rice can be obtained for ₹ 180?
Solution:
Given The Price Of 1 Kg Of Rice Is ₹ 15.
Let, the price of x kg of rice = ₹ y.
It is given that, the price of 1 kg of rice = ₹ 15
∴ price of x kg of rice = ₹ 15x
∴ y= 15x.
It is the equation of the required graph.
From this equation we get,
| X | 1 | 3 | 7 |
| y | 15 | 45 | 105 |
Let, on the graph paper 3 sides of the smallest square on the x-axis represent 1 kg and 1 side of the smallest square on the y-axis represent ₹ 5. The graph OP of the straight line y = 15x is drawn. From the graph, it is found that the abscissa of point Q is 5 units, and its ordinate = ₹ 75 units. Hence, the price of 5 kg of rice is ₹ 75. Again, the ordinate of the point R is 180 units, and its abscissa = 12 units.
Hence, 12 kg of rice can be obtained for ₹180.
Key Terms Related to Graphs
Example 2
If the speed of a man is 5 km per hour, calculate from the graph:
1. How far will he travel in 5 hours?
2. At what time will he travel 40 km?
Solution :
Given The Speed Of A Man Is 5 Km Per Hour.
Let, the man travels y km in x hours. It is given that,
In 1 hour the man travels 5 km
∴ In x hours the man travels 5x km.
y = 5x.
It is the equation of the required graph.
From this equation we get,
| X | 1 | 3 | 7 |
| y | 5 | 15 | 35 |
Let, on the graph paper 5 sides of the smallest square on the x-axis represent 1 hour and 1 side of the smallest square on the y-axis represent a distance of 1 km. The graph OP of the straight line y = 5x is drawn. From the graph, it is found that the abscissa of point Q is 5 units, and its ordinate = 25 units.
In 5 hours the man travels 25 km. Again, the ordinate of the point R is 40 units, and its abscissa = 8 units.
∴ To travel 40 km the man will take 8 hours.
Conceptual Questions on Graphs and Functions
Example 3
The distance of house of Anwar’s maternal uncle is 38 km from the house of Anwar. Anwar starts on a bicycle at 7 a.m. After 2 1/2 hours ride, the bicycle develops trouble. He stops for half an hour to repair it but is not successful. So he is to travel the remaining distance on foot. If the speed of cycling and walking are 12 km and 4 km per hour respectively, when does he reach the house of his maternal uncle? (Solve graphically)
Solution:
Given:-
The distance of house of Anwar’s maternal uncle is 38 km from the house of Anwar.
Anwar starts on a bicycle at 7 a.m. After 2 1/2 hours ride, the bicycle develops trouble.
He stops for half an hour to repair it but is not successful.
So he is to travel the remaining distance on foot.
If the speed of cycling and walking are 12 km and 4 km per hour respectively.
Let, on the graph paper 5 sides of the smallest square on the x-axis represent a distance of 1 km.
Anwar starts from the origin at 7 a.m. and travels by bicycle at a speed of 12 km per hour. Then after 1 hour he travels a distance of 12 km and reaches point A. After this in one more hour of traveling
12 km more he reaches point B. After
this, in 1/2 hour more traveling 6 km more he reaches point C.
After this, he stops for half an hour to repair the bicycle but in vain. The rest period of half an hour has been shown by CD.
After this Anwar starts walking at a speed of 4 km per hour and reaches E after 1 hour and then in one hour more walking 4 km he reaches his maternal uncle’s house at F. Point F is at a distance of 38 km from the starting point. The time at which he reaches point F is found in the graph. It is 5 hours after the start i.e., 12 noon.
∴ Anwar reached the house of their maternal uncle at 12 noon.
Example 4
Abani starts at 10 a.m. at the rate of 5 km per hour. two hours later Biswajit follows him by bicycle at the rate of 7 km per hour. Find graphically when and where they meet each other.
Solution:
Abani Starts At 10 A.M. At The Rate Of 5 Km Per Hour.
Two Hours Later Biswajit Follows Him By Bicycle At The Rate Of 7 Km Per Hour.
Given:-
Let, on the graph paper 5 sides of the smallest square on the x-axis represent 1 hour and 1 side of the smallest square on the y-axis represent a distance f 1 km. Let, Abani starts from the origin O and travels at a speed of 5 km per hour.
Let, the coordinates of point A be (1, 5). Therefore, the abscissa of point A is 1 hour and its ordinate represents 5 km. Hence point A is a point on the motion graph of Abani.
Again Biswajit starts 2 hours after Abani starts. Hence in those two hours, Biswajit has not traveled. That means the distance traveled by Biswajit in two hours is zero. After this, Biswajit starts walking at the rate of 7 km per hour. Now consider point (2, 0) as origin the of point B is plotted whose coordinates are (1, 7). Now joining point B with the point (2, 0) obtained by considering 0 as the origin and producing it we obtain the motion graph of Biswajit for the time after 2 hours.
The two motion graphs intersect each other at point C. Taking O as the origin the coordinates of point C is (7 35)
So 7 hours after Abani starts, they will meet at 5 p.m. at a distance of 35 km from the starting point.
Examples of Real-Life Applications of Graphs
Example 5
The total expenditure of a hostel is partly constant and partly proportional to the number of boarders. If the number of boards is 50, the total expenditure is 6000, and if the number of boards is 70, the total expenditure is 8000. Calculate graphically the total expenditure if the number of boards is 90. Also, find out the constant part of the expenditure.
Solution:
Given:-
The total expenditure of a hostel is partly constant and partly proportional to the number of boarders.
If the number of boards is 50, the total expenditure is 6000, and if the number of boards is 70, the total expenditure is 8000.
A graph is plotted using the given data. The scale chosen is as follows :
1 small div. of x-axis = 2 borders
5 small div. of y-axis = ₹ 1000
Point A corresponds to a total expenditure of ₹ 6000 for 50 borders and point B corresponds to a total expenditure of ₹ 8000 for 70 borders. The straight line joining the two points is drawn.
Point P on the straight line represents the expenditure of the hostel when the number of boarders is 90. From the graph, it is found to be ^ 10,000.
The line AB intersects the y-axis at (0,1000).
∴ The fixed component of the expenditure = ₹ 1000.
The total expenditure is ₹ 10,000
when the number of borders is 90 and the constant or fixed part of the expenditure is ₹ 1000.
∴ The total expenditure is ₹ 10,000 when the number of borders is 90 and the constant or fixed part of the expenditure is ₹ 1000.