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Coordinate Geometry - Cartesian System and Plotting of Points

Grade 9ICSE

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

🔑Concepts

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The Cartesian Plane: It is a two-dimensional surface formed by the intersection of two perpendicular number lines. The horizontal line is called the xx-axis (represented as X′OXX'OX), and the vertical line is called the yy-axis (represented as Y′OYY'OY). These axes divide the plane into four infinite regions.

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The Origin: The point where the xx-axis and yy-axis intersect is called the origin, denoted by the letter OO. Its coordinates are always (0,0)(0, 0), serving as the starting point for measuring distances in the system.

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Coordinates of a Point: Every point in the plane is identified by an ordered pair of numbers (x,y)(x, y). The first number, xx, is the 'Abscissa' (perpendicular distance from the yy-axis), and the second number, yy, is the 'Ordinate' (perpendicular distance from the xx-axis).

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Quadrants and Sign Convention: The axes divide the plane into four Quadrants, numbered I to IV counter-clockwise. In Quadrant I (top-right), signs are (+,+)(+, +); in Quadrant II (top-left), signs are (−,+)(-, +); in Quadrant III (bottom-left), signs are (−,−)(-,-); and in Quadrant IV (bottom-right), signs are (+,−)(+, -). Any point on the axes does not belong to any quadrant.

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Points on the Axes: If a point lies on the xx-axis, its distance from the xx-axis is zero, so its coordinates are of the form (x,0)(x, 0). Conversely, if a point lies on the yy-axis, its distance from the yy-axis is zero, so its coordinates are of the form (0,y)(0, y). Identifying these is crucial for plotting intercepts.

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Plotting a Point: To plot a point P(a,b)P(a, b), start at the origin (0,0)(0, 0). Move ∣a∣|a| units along the xx-axis (right if a>0a > 0, left if a<0a < 0). From that position, move ∣b∣|b| units parallel to the yy-axis (up if b>0b > 0, down if b<0b < 0). The final position is the location of the point.

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Distance from Axes: For any point P(x,y)P(x, y), the perpendicular distance from the xx-axis is equal to the absolute value of the yy-coordinate, i.e., ∣y∣|y|. The perpendicular distance from the yy-axis is equal to the absolute value of the xx-coordinate, i.e., ∣x∣|x|.

📐Formulae

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

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

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

General form of a point on xx-axis: (x,0)(x, 0)

General form of a point on yy-axis: (0,y)(0, y)

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:

Identify the quadrant or axis for the following points without plotting them: A(5,−3)A(5, -3), B(−2,−7)B(-2, -7), C(0,4)C(0, 4), and D(−3,8)D(-3, 8).

Solution:

  1. For point A(5,−3)A(5, -3): The xx-coordinate is positive and the yy-coordinate is negative (+,−+, -). This corresponds to Quadrant IV.
  2. For point B(−2,−7)B(-2, -7): Both xx and yy coordinates are negative (−,−-, -). This corresponds to Quadrant III.
  3. For point C(0,4)C(0, 4): The xx-coordinate is 00. Any point with x=0x = 0 lies on the yy-axis. Since yy is positive, it is on the positive yy-axis.
  4. For point D(−3,8)D(-3, 8): The xx-coordinate is negative and the yy-coordinate is positive (−,+-, +). This corresponds to Quadrant II.

Explanation:

Quadrants are determined by the signs of the coordinates: I (+,+)(+,+), II (−,+)(-,+), III (−,−)(-,-), and IV (+,−)(+,-). Points with a zero coordinate lie on the axes.

Problem 2:

Find the distance between the points P(−3,4)P(-3, 4) and Q(5,−2)Q(5, -2).

Solution:

Step 1: Identify coordinates: (x1,y1)=(−3,4)(x_1, y_1) = (-3, 4) and (x2,y2)=(5,−2)(x_2, y_2) = (5, -2). Step 2: Use the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Step 3: Substitute values: d=(5−(−3))2+(−2−4)2d = \sqrt{(5 - (-3))^2 + (-2 - 4)^2}. Step 4: Simplify inside the square root: d=(5+3)2+(−6)2=82+(−6)2d = \sqrt{(5 + 3)^2 + (-6)^2} = \sqrt{8^2 + (-6)^2}. Step 5: Calculate squares: d=64+36=100d = \sqrt{64 + 36} = \sqrt{100}. Step 6: Final result: d=10d = 10 units.

Explanation:

The distance formula is derived from the Pythagoras theorem applied to the horizontal and vertical distances between two points.