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Introduction to Graphs - The Cartesian Plane and Coordinates

Grade 8ICSE

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

🔑Concepts

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The Cartesian plane is formed by two perpendicular number lines: the horizontal xx-axis and the vertical yy-axis. Their point of intersection is the origin O(0,0)O(0, 0). Any point PP is represented by an ordered pair (x,y)(x, y), where xx is the abscissa (distance from yy-axis) and yy is the ordinate (distance from xx-axis).

A Cartesian plane showing axes, origin, and the point P(3, 2).
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The axes divide the plane into four regions called quadrants. In Quadrant I, both xx and yy are positive (+,+)(+, +). In Quadrant II, xx is negative and yy is positive (−,+)(-, +). In Quadrant III, both are negative (−,−)(-, -). In Quadrant IV, xx is positive and yy is negative (+,−)(+, -). Points on the axes do not belong to any quadrant.

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A linear equation in two variables, such as y=x+1y = x + 1, can be represented as a straight line on the Cartesian plane. Every point (x,y)(x, y) that lies on this line satisfies the given equation.

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Special cases of lines include vertical lines (x=kx = k) which are parallel to the yy-axis, and horizontal lines (y=ky = k) which are parallel to the xx-axis.

📐Formulae

General representation of a point: P(x,y)P(x, y)

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

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

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

General form of a linear equation: y=mx+cy = mx + c

💡Examples

Problem 1:

Identify the quadrant or axis for the following points without plotting them: A(3,4)A(3, 4), B(−2,5)B(-2, 5), C(0,−3)C(0, -3), and D(4,−2)D(4, -2).

Solution:

  1. For A(3,4)A(3, 4): Both xx and yy are positive (+,+)(+, +), so it lies in Quadrant I.
  2. For B(−2,5)B(-2, 5): xx is negative and yy is positive (−,+)(-, +), so it lies in Quadrant II.
  3. For C(0,−3)C(0, -3): The xx-coordinate is 00. Any point with x=0x=0 lies on the yy-axis.
  4. For D(4,−2)D(4, -2): xx is positive and yy is negative (+,−)(+, -), so it lies in Quadrant IV.

Explanation:

The location of a point is determined by the signs of its coordinates. If one coordinate is zero, the point lies on an axis rather than in a quadrant.

Problem 2:

Given the equation y=2x−1y = 2x - 1, find the coordinates of the points where the line crosses the xx-axis and the yy-axis.

Solution:

  1. To find the yy-axis intersection (where x=0x = 0): y=2(0)−1y = 2(0) - 1 y=−1y = -1 So, the point is (0,−1)(0, -1).

  2. To find the xx-axis intersection (where y=0y = 0): 0=2x−10 = 2x - 1 2x=12x = 1 x=frac12=0.5x = \\frac{1}{2} = 0.5 So, the point is (0.5,0)(0.5, 0).

Explanation:

Intersection with the yy-axis always occurs when x=0x=0, and intersection with the xx-axis always occurs when y=0y=0. We substitute these values into the linear equation to solve for the unknown coordinate.

Problem 3:

Plot the points A(2,3)A(2, 3), B(−3,3)B(-3, 3), C(−3,−2)C(-3, -2), and D(2,−2)D(2, -2) on a Cartesian plane. Join them in order A→B→C→D→AA \rightarrow B \rightarrow C \rightarrow D \rightarrow A. Name the geometrical figure formed and calculate its area.

A square ABCD plotted on a coordinate grid with vertices at (2,3), (-3,3), (-3,-2), and (2,-2).

Solution:

  1. Plotting the points:
  • A(2,3)A(2, 3) is in Quadrant I.
  • B(−3,3)B(-3, 3) is in Quadrant II.
  • C(−3,−2)C(-3, -2) is in Quadrant III.
  • D(2,−2)D(2, -2) is in Quadrant IV.

  1. Joining the points forms a rectangle ABCDABCD.

  2. Calculating dimensions:

  • Length AB=∣2−(−3)∣=5AB = |2 - (-3)| = 5 units.
  • Breadth BC=∣3−(−2)∣=5BC = |3 - (-2)| = 5 units.
  1. Since all sides are equal and angles are 90∘90^{\circ}, the figure is a square. Area = side×side=5×5=25side \times side = 5 \times 5 = 25 sq. units.

Explanation:

By plotting the given coordinates and connecting them, we observe a closed quadrilateral. Since the horizontal distance (5 units) and vertical distance (5 units) between vertices are equal, the figure is a square.

Problem 4:

Draw the graph of the function y=x+2y = x + 2 by finding at least three points that satisfy the equation.

A straight line graph of the equation y = x + 2 passing through points (-2, 0), (0, 2), and (2, 4).

Solution:

  1. Create a table of values:
  • If x=0x = 0, y=0+2=2y = 0 + 2 = 2. Point is (0,2)(0, 2).
  • If x=2x = 2, y=2+2=4y = 2 + 2 = 4. Point is (2,4)(2, 4).
  • If x=−2x = -2, y=−2+2=0y = -2 + 2 = 0. Point is (−2,0)(-2, 0).
  1. Plot the points (0,2)(0, 2), (2,4)(2, 4), and (−2,0)(-2, 0) on the Cartesian plane.
  2. Draw a straight line passing through these points.

Explanation:

To graph a linear equation, we select arbitrary values for xx, calculate the corresponding yy values, plot the resulting ordered pairs, and join them with a straight line.