These lie on a line. We can name it as KL or KM or MN etc. It is parallel to x-axis.
These lie on a line. We can name it as XY or WY or YZ etc.
Note that in each of the above cases, graph obtained by joining the plotted points is a line. Such graphs are called linear graphs.

0 to 5 min and 5 to 10 min

The line graph for the area of a square as per the given table.

No, it is not a linear graph because when we join the points, then there is not a straight line formed.

The histogram of given data is shown below.

 

A bar graph for the given temperature in month. from March to July.

Given points are A(1, 2), 8(3,4) and C(5. 2).

Location of given points on the graph given below So, the diagonals of this rhombus are equal in length. because this is a square, a kind of rhombus, whose diagonals are same.

The histogram of given data is shown below.

Given. y-coordinate = 2 times of x-coordinate.
If x = 2, then y = 2x = 2 x 2 = 4
If x = 3, then y = 2x = 2 x 3 = 6 

So, the points are (2, 4) and (3. 6) and graph is shown below:

To draw the graph for the given table of values, we use the following steps:
Step I:  Draw X-axis and Y-axis and take scale,on horizontal axis 3 units = 1h, on vertical axis 1 unit = 20 km
Step II : Mark the time (in hours) on horizontal axis and distance (in Ian) on vertical axis.
Step III : Now, plot the points (6, 40), (7, 80), (8, 120) and (9, 160) on graph paper.
Step IV : Join these points.
Then, we get a linear graph which is shown below

(i) In the graph, corresponding to 7:30 am on horizontal axis,we get the distance 100 Ian on vertical axis i.e. point P. Distance covered during the period 7:30 am to 8: 00 am = (120 -100) km = 20 km
(ii) In the given graph, ( from point P) at 7: 30 am the car had covered'a distance of 100 km, since it's start.

According to the given information, the table is as follows

We get a table of values.

(i) Scale: Horizontal: 2 units = 1 h; Vertical: 1 unit = 10 km
(ii) Mark time on horizontal axis.
(iii) Mark distance on vertical axis.
(iv) Plot the points (1, 60), (2, 120), (3, 180), (4, 240)
(a)\(1\frac { 1 }{ 4 } \) or 1 h 15 min
(b) 210 km

According to the given information, the table is as follows:

Steps to follow
Step I Find the quantities to be plotted as deposit and SI.
Step II Decide the quantities to be taken on X-axis and on Y-axis.
Step III Choose a scale.
Step IV Plot points.
Step V Join the points.
We get a table of values:

(i) Scale: 1 unit = Rs. 100 on horizontal axis, 1 unit = Rs 20 on vertical axis.
(ii) Mark deposits along horizontal axis,
(iii) Mark simple interest along vertical axis,
(iv) Plot the points: (100,20), (200,40), (300,60), (500, 100), etc.
(v) Join the points. We get a graph, i.e a line.

(a) Corresponding to Rs 450 on horizontal axis, we get the interest to be Rs. 90 on vertical axis.
(b) Corresponding to Rs 90 on the vertical axis, we get the sum to be Rs. 450 on the horizontal axis.

Plotting of points (4, 2) (2,4) (3, 3) (5,4).

No, the points do not lie on a straight line. These points form a triangle.

Total number of tourists = 1850
Difference between highest arrival and lowest arrival = 350

We know that, if time is increasing, then-temperature will either be increase or decrease or constant.
(i) Yes, it is the time-temperature graph, because it shows increase in temperature with increase in time.
(ii) Yes, it is the time-temperature graph, because it shows decrease in temperature with increase in time.
(iii) No, it is not the time-temperature graph, because it shows that at a particular time-temperature increase (i.e. different temperature) which is not possible.
(iv) Yes, it is the time-temperature graph, because it shows that at different time (or increasing time) temperature is constant (or same).

(a) In the given graph, the scale is taken for the time axis (i.e. along X-axis) is 4 units = 1h.
(b) The person starts travel at 8 am and end at 11: 30 am.
So, total travelling time = Difference between 8 am to 11: 30 am = \(3\frac{1}{2}\)h or 3 h 30 min
(c) From the given graph, it is clear that the distance of place of the merchant from the town is 22 km.
(d) Yes, from the given graph, it is clear that the person was 16 km away from town when the time was 10 am to 10: 30 am. This shows that the person did not tray during the interval 10 am to 10: 30 am. The horizons line segment representing travel during this period is illustrative of this fact.
(e) Distance covered by person between 8 am to 9 arn = 10krn
Distance covered by person between 9 am to 10 am = (16 - 10)km = 6 km
Distance covered by person between 10 am to 10: 30 am = (16 -16) km = 0 km
Distance covered by person between 10: 30 am to 11: 30 am = (22 -16) km = 6 km
Hence, we can say that the courier person ride fastestin between 8 am to 9 am.

Firstly, draw two perpendicular lines on the graph paper and then take year along X-axis (horizontal line) and population (in thousand) along Y-axis (vertical line).
Now, mark the points (2003, 12), (2004, 12.5), (2005, 13), (2006, 13.2) and (2007, 13.5) on the graph paper and join them by dark line. Thus, we get linear graph for men. Again, mark the points (2003, 11.3), (2004, 11.9), (2005, 13), (2006, 13.6) and (2007, 12.8) on the graph paper and join them by dotted lines. Thus, we get linear graph for women. Hence, the required linear graph for population (in thousands) of men and women in a village in different years is given below.

Firstly, draw two perpendicular lines on the graph paper and then take year along X-axis and days along Y-axis. Now, mark the points (2003, 8), (2004, 10), (2005, 5) and (2006, 12) on the graph paper and join them. Thus, we get a linear graph which shows snowfall in different years.

(a) From the given graph, it is clear that the patient's temperature at 1 pm was36.5°C.
(b) From the given graph, it is clear that the patient's temperature was 38.5°C at 12 noon.
(c) From the given graph, it is clear that the patient's temperature was same i.e. 36.5°C at 1 pm and 2 pm.
(d) From the given graph, it is clear that the patient's temperature at 1 pm and 2 pm was same, which is 36.5°C. This shows that temperature did not change during the interval 1 pm to 2 pm. Also, 1: 30 pm is between 1 pm and 2 pm, so the temperature at 1: 30 pm was 36.5°C.
(e) From the given graph, it is clear that the patient's temperature showed an upward trend during the periods 9 am to 10 am, 10 am to 11 am and 2 pm to 3 pm, because temperature is increasing in these intervals.

Percentage of houses using solar energy = \(\frac{10}{36}\times 100=\frac{1000}{36}\) = 27.77%
Percentage of houses using gas energy = \(\frac{8}{36}\times 100\) = 22.22%
Percentage of houses using electricity energy = \(\frac{12}{36}\times 100=\frac{1200}{36}\) = 33.33%
Percentage of houses using oil energy = \(\frac{6}{36}\times\) 100 = 16.66%
Pie graph of the above data is given below:

(i) Let us take a suitable scale on both the axes (Fig)

(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.