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Introduction to Graphs - Linear Graphs

Grade 8ICSE

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

🔑Concepts

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The Cartesian Coordinate System: A graph is formed by two perpendicular number lines. The horizontal line is the xx-axis, and the vertical line is the yy-axis. Their intersection point is the Origin (0,0)(0, 0). Any point PP is represented by an ordered pair (x,y)(x, y), where xx is the abscissa and yy is the ordinate.

Cartesian plane showing point P at coordinates (3, 2) and the Origin.
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Linear Graphs: A graph that is a single straight line is called a linear graph. It represents a relationship where a change in one variable results in a proportional change in the other. Equations like y=mxy = mx result in lines passing through the origin.

A linear graph of the equation y = x passing through the origin.
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Plotting Points: To draw a linear graph, we first create a table of values by choosing independent values for xx and calculating the corresponding yy values. At least two points are needed to draw a unique straight line, though three are preferred for accuracy.

Multiple points plotted to form a straight line.
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Horizontal and Vertical Lines: An equation of the form x=kx = k represents a vertical line parallel to the yy-axis. An equation of the form y=ky = k represents a horizontal line parallel to the xx-axis.

Graph showing a horizontal line y=3 and a vertical line x=-2.

📐Formulae

General form of a linear equation: ax+by+c=0ax + by + c = 0

Slope-intercept form: y=mx+cy = mx + c

Equation of the xx-axis: y=0y = 0

Equation of the yy-axis: x=0x = 0

Coordinates of the Origin: (0,0)(0, 0)

💡Examples

Problem 1:

Draw a linear graph for the equation y=x+2y = x + 2.

Solution:

  1. Create a table of values by choosing arbitrary values for xx:
  • If x=0x = 0, y=0+2=2y = 0 + 2 = 2. Point is (0,2)(0, 2).
  • If x=1x = 1, y=1+2=3y = 1 + 2 = 3. Point is (1,3)(1, 3).
  • If x=−2x = -2, y=−2+2=0y = -2 + 2 = 0. Point is (−2,0)(-2, 0).
  1. Plot these three points (0,2)(0, 2), (1,3)(1, 3), and (−2,0)(-2, 0) on the Cartesian plane.
  2. Using a ruler, draw a straight line passing through all these points.
  3. Label the line as y=x+2y = x + 2.

Explanation:

To graph any linear equation, find at least two or three points that satisfy the equation, plot them, and join them with a straight line. Using three points ensures accuracy.

Problem 2:

Find the points where the line 3x+2y=123x + 2y = 12 intersects the xx-axis and the yy-axis.

Solution:

  1. To find the xx-intercept, set y=0y = 0: 3x+2(0)=123x + 2(0) = 12 3x=123x = 12 x=frac123=4x = \\frac{12}{3} = 4 The line intersects the xx-axis at (4,0)(4, 0).

  2. To find the yy-intercept, set x=0x = 0: 3(0)+2y=123(0) + 2y = 12 2y=122y = 12 y=frac122=6y = \\frac{12}{2} = 6 The line intersects the yy-axis at (0,6)(0, 6).

Explanation:

Intersects are found by setting one coordinate to zero. The xx-intercept occurs when the vertical distance is zero, and the yy-intercept occurs when the horizontal distance is zero.

Problem 3:

A car travels at a constant speed of 40 km/h40 \text{ km/h}. Draw a distance-time graph for this motion and find the distance covered in 3.53.5 hours.

Distance-time graph showing a straight line representing 40 km/h.

Solution:

  1. Let xx represent time in hours and yy represent distance in km.
  2. The relationship is given by Distance = Speed ×\times Time, so y=40xy = 40x.
  3. Table of values:
  • x=1,y=40x = 1, y = 40
  • x=2,y=80x = 2, y = 80
  • x=3,y=120x = 3, y = 120
  1. Plot the points (1,40),(2,80),(3,120)(1, 40), (2, 80), (3, 120) and join them to the origin (0,0)(0, 0).
  2. For x=3.5x = 3.5, the graph shows y=140y = 140.

Explanation:

The distance-time graph for constant speed is always a straight line starting from the origin. The slope represents the speed of the car.

Problem 4:

A bank offers 5%5\% simple interest per annum on deposits. Draw a graph to show the relation between the deposit amount (xx) and the interest earned (yy) in one year for deposits up to Rs 50005000. From the graph, find the interest for a deposit of Rs 35003500.

A linear graph showing interest y vs deposit x, where a point is marked at x=3500, y=175.

Solution:

  1. Establish the Relation: The simple interest II for one year is given by I=P×R×T100I = \frac{P \times R \times T}{100}. Here, R=5R = 5, T=1T = 1, and PP is the deposit xx. So, y=x×5×1100⇒y=0.05xy = \frac{x \times 5 \times 1}{100} \Rightarrow y = 0.05x or y=x20y = \frac{x}{20}.

  2. Generate Data Points:

  • If x=1000x = 1000, y=100020=50y = \frac{1000}{20} = 50
  • If x=2000x = 2000, y=200020=100y = \frac{2000}{20} = 100
  • If x=3000x = 3000, y=300020=150y = \frac{3000}{20} = 150
  • If x=4000x = 4000, y=400020=200y = \frac{4000}{20} = 200
  • If x=5000x = 5000, y=500020=250y = \frac{5000}{20} = 250

  1. Plotting the Graph: Plot the points (1000,50),(2000,100),…,(5000,250)(1000, 50), (2000, 100), \dots, (5000, 250) and join them with a straight line passing through the origin (0,0)(0, 0).

  2. Finding Interest for Rs 3500: On the graph, locate x=3500x = 3500 on the x-axis. Move vertically to meet the line and then horizontally to the y-axis. The value on the y-axis is 175175.

Final Answer: The interest for a deposit of Rs 35003500 is Rs 175175.

Explanation:

The relationship between principal and simple interest for a fixed rate and time is a direct proportion, resulting in a linear graph of the form y=mxy = mx. The constant of proportionality (slope) here is 0.050.05.