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Introduction to Graphs - Line Graphs

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A line graph consists of bits of line segments joined consecutively which displays data that changes continuously over periods of time. It uses a coordinate plane with a horizontal axis (xx-axis) and a vertical axis (yy-axis).

A basic line graph on a coordinate plane showing segments connecting discrete points.
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The xx-coordinate (abscissa) represents the independent variable (like time), and the yy-coordinate (ordinate) represents the dependent variable (like temperature or distance).

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A 'Linear Graph' is a special case of a line graph where all the points lie on a single straight line, indicating a constant rate of change between variables.

A straight line graph passing through the origin representing a linear relationship.
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A kink (zigzag line) is used on an axis if the data values do not start from zero, allowing for a clearer representation of the range where data actually exists.

📐Formulae

Coordinates of a point: (x,y)(x, y) where xx is the abscissa and yy is the ordinate.

General equation of a straight line (Linear Graph): y=mx+cy = mx + c, where mm is the slope and cc is the yy-intercept.

Slope (mm) between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2): m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

Midpoint Formula: M=(x1+x22,y1+y22)M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

Distance between two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2): d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

💡Examples

Problem 1:

A car travels at a constant speed of 40 km/h40\text{ km/h}. Create a distance-time table for 0,1,2,30, 1, 2, 3 hours and describe how to plot the resulting linear graph.

Solution:

Step 1: Create the table based on the formula Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}.

  • At t=0 h,d=40×0=0 kmt = 0\text{ h}, d = 40 \times 0 = 0\text{ km}. Point: (0,0)(0, 0)
  • At t=1 h,d=40×1=40 kmt = 1\text{ h}, d = 40 \times 1 = 40\text{ km}. Point: (1,40)(1, 40)
  • At t=2 h,d=40×2=80 kmt = 2\text{ h}, d = 40 \times 2 = 80\text{ km}. Point: (2,80)(2, 80)
  • At t=3 h,d=40×3=120 kmt = 3\text{ h}, d = 40 \times 3 = 120\text{ km}. Point: (3,120)(3, 120)

Step 2: On a graph paper, take Time on the xx-axis (1 unit=1 hour1\text{ unit} = 1\text{ hour}) and Distance on the yy-axis (1 unit=20 km1\text{ unit} = 20\text{ km}). Step 3: Plot the points (0,0),(1,40),(2,80),(3,120)(0,0), (1,40), (2,80), (3,120) and join them with a straight line.

Explanation:

Since the speed is constant, the relationship between distance and time is linear. The graph will be a straight line passing through the origin, showing that distance is directly proportional to time.

Problem 2:

The following coordinates represent the temperature recorded at different times of the day: (6 AM,15∘C),(9 AM,20∘C),(12 PM,25∘C),(3 PM,22∘C)(6\text{ AM}, 15^{\circ}\text{C}), (9\text{ AM}, 20^{\circ}\text{C}), (12\text{ PM}, 25^{\circ}\text{C}), (3\text{ PM}, 22^{\circ}\text{C}). Find the increase in temperature between 6 AM6\text{ AM} and 12 PM12\text{ PM}.

Solution:

Step 1: Identify the yy-coordinates (temperatures) for the given times.

  • Temperature at 6 AM(y1)=15∘C6\text{ AM} (y_1) = 15^{\circ}\text{C}
  • Temperature at 12 PM(y2)=25∘C12\text{ PM} (y_2) = 25^{\circ}\text{C}

Step 2: Calculate the difference (increase). Increase=y2−y1\text{Increase} = y_2 - y_1 Increase=25∘C−15∘C=10∘C\text{Increase} = 25^{\circ}\text{C} - 15^{\circ}\text{C} = 10^{\circ}\text{C}

Explanation:

By reading the yy-values corresponding to the specific points on the xx-axis (time), we can determine the change in the dependent variable (temperature) over the specified interval.

Problem 3:

Plot a line graph for the following data showing the side of a square and its perimeter: Side (in cm): 2,3,4,52, 3, 4, 5. Perimeter (in cm): 8,12,16,208, 12, 16, 20. Determine if it is a linear graph.

Linear graph showing the relationship between square side and perimeter.

Solution:

  1. Choose a scale: Let 11 unit on xx-axis =1 cm= 1\text{ cm} and 11 unit on yy-axis =4 cm= 4\text{ cm}.
  2. Plot points: (2,8),(3,12),(4,16),(5,20)(2, 8), (3, 12), (4, 16), (5, 20).
  3. Join the points with line segments. Since all points lie on a single straight line, it is a linear graph.

Explanation:

Because the perimeter PP is related to side ss by P=4sP = 4s, the ratio is constant, resulting in a straight line through the origin.

Problem 4:

A patient's temperature was recorded every hour. Plot the data: 9 AM:37∘C,10 AM:40∘C,11 AM:39∘C,12 PM:38∘C9\text{ AM}: 37^{\circ}\text{C}, 10\text{ AM}: 40^{\circ}\text{C}, 11\text{ AM}: 39^{\circ}\text{C}, 12\text{ PM}: 38^{\circ}\text{C}.

A line graph showing temperature fluctuations over four hours.

Solution:

  1. Let xx-axis represent time (hours) and yy-axis represent temperature (∘C^{\circ}\text{C}).
  2. Plot (9,37),(10,40),(11,39),(12,38)(9, 37), (10, 40), (11, 39), (12, 38).
  3. Connect points with straight line segments.

Explanation:

This graph shows the fluctuation of body temperature over time. It is a line graph but not a linear graph because the rate of change is not constant.