A line graph consists of bits of line segments joined consecutively. Sometimes the graph may be a whole unbroken line. Such a graph is called a linear graph. To draw such a line we need to locate some points on the graph sheet. We will now learn how to locate points conveniently on a graph sheet.
The teacher put a dot on the black-board. She asked the students how they would describe its location. There were several responses
Can any one of these statements help fix the position of the dot? No! Why not? Think about it.
John then gave a suggestion. He measured the distance of the dot from the left edge of the board and said, “The dot is 90 cm from the left edge of the board”. Do you think John’s suggestion is really helpful? (Fig 15.10)
A, A1, A2, A3 are all 90 cm away A is 90 cm from left edge and
from the left edge. 160 cm from the bottom edge
Rekha then came up with a modified statement : “The dot is 90 cm from the left edge and 160 cm from the bottom edge”. That solved the problem completely! (Fig 15.11) The teacher said, “We describe the position of this dot by writing it as (90, 160)”. Will the point (160, 90) be different from (90, 160)? Think about it.
The 17th century mathematician Rene Descartes, it is said, noticed the movement of an insect near a corner of the ceiling and began to think of determining the position of a given point in a plane. His system of fixing a point with the help of two measurements, vertical and horizontal, came to be known as Cartesian system, in his honour
Suppose you go to an auditorium and search for your reserved seat. You need to know two numbers, the row number and the seat number. This is the basic method for fixing a point in a plane.
Observe in Fig 15.12 how the point (3, 4) which is 3 units from left edge and 4 units from bottom edge is plotted on a graph sheet. The graph sheet itself is a square grid. We draw the x and y axes conveniently and
then fix the required point. 3 is called the x-coordinate of the point; 4 is the y-coordinate of the point. We say that the coordinates of the point are (3, 4).
Example 3:Plot the point (4, 3) on a graph sheet. Is it the same as the point (3, 4)? Solution: Locate the x, y axes, (they are actually number lines!). Start at O (0, 0). Move 4 units to the right; then move 3 units up, you reach the point (4, 3). From Fig 15.13, you can see that the points (3, 4) and (4, 3) are two different points. Example 4: From Fig 15.14, choose the letter(s) that indicate the location of the points given below:
(i) (2, 1)
(ii) (0, 5)
(iii) (2, 0)
(iv) The coordinates of A.
(v) The coordinates of F.
(i) (2, 1) is the point E (It is not D!).
(ii) (0, 5) is the point B (why? Discuss with your friends!). (iii) (2, 0) is the point G.
(iv) Point A is (4, 5) (v) F is (5.5, 0)
Example 5: Plot the following points and verify if they lie on a line. If they lie on a line,name it.
(i) (0, 2), (0, 5), (0, 6), (0, 3.5) (ii) A (1, 1), B (1, 2), C (1, 3), D (1, 4)
(iii) K (1, 3), L (2, 3), M (3, 3), N (4, 3) (iv) W (2, 6), X (3, 5), Y (5, 3), Z (6, 2)
Note that in each of the above cases, graph obtained by joining the plotted points is a line. Such graphs are called linear graphs.
In everyday life, you might have observed that the more you use a facility, the more you pay for it. If more electricity is consumed, the bill is bound to be high. If less electricity is used, then the bill will be easily manageable. This is an instance where one quantity affects another. Amount of electric bill depends on the quantity of electricity used. We say that the quantity of electricity is an independent variable (or sometimes control variable) and the amount of electric bill is the dependent variable. The relation between such variables can be shown through a graph.
The following table gives the quantity of petrol and its cost.
|No. of Litres of petrol||10||15||20||25|
|Cost of petrol in Rs||500||750||1000||1250|
Plot a graph to show the data.
Solution: (i) Let us take a suitable scale on both the axes (Fig 15.16).
(ii) Mark number of litres along the horizontal axis.
(iii) Mark cost of petrol along the vertical axis.
(iv) Plot the points: (10,500), (15,750), (20,1000), (25,1250).
(v) Join the points.
We find that the graph is a line. (It is a linear graph). Why does this graph pass through the origin? Think about it.
This graph can help us to estimate a few things. Suppose we want to find the amount needed to buy 12 litres of petrol. Locate 12 on the horizontal axis. Follow the vertical line through 12 till you meet the graph at P (say).
From P you take a horizontal line to meet the vertical axis. This meeting point provides the answer.
This is the graph of a situation in which two quantities, are in direct variation. (How ?). In such situations, the graphs will always be linear.
Example: (Principal and Simple Interest)
A bank gives 10% Simple Interest (S.I.) on deposits by senior citizens. Draw a graph to illustrate the relation between the sum deposited and simple interest earned. Find from your graph
(a) the annual interest obtainable for an investment of Rs 250.
(b) the investment one has to make to get an annual simple interest of Rs 70.
(i) Scale : 1 unit = Rs 100 on horizontal axis; 1 unit = Rs 10 on vertical axis.
(ii) Mark Deposits along horizontal axis.
(iii) Mark Simple Interest along vertical axis.
(iv) Plot the points : (100,10), (200, 20), (300, 30), (500,50) etc.
(v) Join the points. We get a graph that is a line (Fig 15.17).
(a) Corresponding to Rs 250 on horizontal axis, we get the interest to be Rs 25 on vertical axis.
(b) Corresponding to Rs 70 on the vertical axis, we get the sum to be Rs 700 on the horizontal axis
Example: (Time and Distance)
Ajit can ride a scooter constantly at a speed of 30 kms/hour. Draw a time-distance graph
for this situation. Use it to find
(i) the time taken by Ajit to ride 75 km. (ii) the distance covered by Ajit in hours
(i) Scale: (Fig 15.18)
Horizontal: 2 units = 1 hour
Vertical: 1 unit = 10 km
(ii) Mark time on horizontal axis.
(iii) Mark distance on vertical axis.
(iv) Plot the points: (1, 30), (2, 60), (3, 90), (4, 120).
(v) Join the points. We get a linear graph.
(a) Corresponding to 75 km on the vertical axis, we get the time to be 2.5 hours on the horizontal axis. Thus 2.5 hours are needed to cover 75 km.
(b) Corresponding to hours on the horizontal axis, the distance covered is 105 km on the vertical axis.
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