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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 coordinate system is formed by two mutually perpendicular lines called axes: the horizontal xx-axis and the vertical yy-axis. Their point of intersection is called the Origin O(0,0)O(0, 0). The axes divide the plane into four regions called quadrants: Quadrant I (top right), Quadrant II (top left), Quadrant III (bottom left), and Quadrant IV (bottom right).

A Cartesian plane showing the four quadrants and signs of coordinates.
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Any point in the plane is represented by an ordered pair (x,y)(x, y), where xx is the abscissa (perpendicular distance from the yy-axis) and yy is the ordinate (perpendicular distance from the xx-axis). To plot (3,2)(3, 2), move 3 units right and 2 units up from the origin.

Diagram showing the abscissa and ordinate of point P(3, 2).
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Points on the xx-axis have an ordinate of 0, taking the form (x,0)(x, 0). Points on the yy-axis have an abscissa of 0, taking the form (0,y)(0, y).

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Reflection of a point: The reflection of P(x,y)P(x, y) in the xx-axis is (x,−y)(x, -y). The reflection in the yy-axis is (−x,y)(-x, y). The reflection in the origin is (−x,−y)(-x, -y).

📐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.

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 graph and identify the geometrical figure formed by joining them in order.

A square plotted on the Cartesian plane with vertices A, B, C, D.

Solution:

  1. Plot AA in Q I, BB in Q II, CC in Q III, and DD in Q IV.
  2. Join ABAB, BCBC, CDCD, and DADA.
  3. Length AB=∣2−(−3)∣=5AB = |2 - (-3)| = 5 units.
  4. Length BC=∣3−(−2)∣=5BC = |3 - (-2)| = 5 units.
  5. Since all sides are equal and adjacent sides are perpendicular, ABCDABCD is a square.

Explanation:

By calculating the distances between adjacent vertices using the horizontal and vertical differences, we find that all sides are equal to 5 units. Since the axes are perpendicular, the shape formed by these horizontal and vertical lines is a square.

Problem 4:

Find the coordinates of the midpoint MM of the line segment joining points L(2,4)L(2, 4) and K(6,0)K(6, 0).

Line segment LK with its midpoint M plotted on the coordinate plane.

Solution:

The midpoint M(x,y)M(x, y) of a segment with endpoints (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by: x=x1+x22=2+62=4x = \frac{x_1 + x_2}{2} = \frac{2 + 6}{2} = 4 y=y1+y22=4+02=2y = \frac{y_1 + y_2}{2} = \frac{4 + 0}{2} = 2 Therefore, M=(4,2)M = (4, 2).

Explanation:

The midpoint is the average of the xx-coordinates and the yy-coordinates of the two endpoints.