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Coordinate Geometry - The Cartesian System-advanced

Grade 9CBSE

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

🔑Concepts

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The Cartesian Plane consists of two mutually perpendicular number lines: the horizontal XX-axis and the vertical YY-axis. Their intersection point is the origin O(0,0)O(0, 0). These axes divide the plane into four regions called quadrants: Quadrant I (+,+)(+,+), Quadrant II (−,+)(-,+), Quadrant III (−,−)(-,-), and Quadrant IV (+,−)(+,-). Points on the axes themselves do not belong to any quadrant.

Cartesian plane showing four quadrants and the origin.
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For any point P(x,y)P(x, y), the xx-coordinate (abscissa) represents the perpendicular distance from the YY-axis, and the yy-coordinate (ordinate) represents the perpendicular distance from the XX-axis. If a point lies on the XX-axis, its ordinate is always 00, i.e., (x,0)(x, 0). If a point lies on the YY-axis, its abscissa is always 00, i.e., (0,y)(0, y).

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Horizontal and vertical distances between points can be calculated by looking at the change in coordinates. For two points with the same ordinate (x1,y)(x_1, y) and (x2,y)(x_2, y), the distance is ∣x2−x1∣|x_2 - x_1|. For two points with the same abscissa (x,y1)(x, y_1) and (x,y2)(x, y_2), the distance is ∣y2−y1∣|y_2 - y_1|.

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Symmetry across axes: The reflection of a point P(x,y)P(x, y) about the XX-axis is (x,−y)(x, -y). The reflection of P(x,y)P(x, y) about the YY-axis is (−x,y)(-x, y). The reflection about the origin is (−x,−y)(-x, -y).

📐Formulae

P(x,y)→x=Abscissa,y=OrdinateP(x, y) \rightarrow x = \text{Abscissa}, y = \text{Ordinate}

Distance of P(x,y) from X-axis=∣y∣ units\text{Distance of } P(x, y) \text{ from } X\text{-axis} = |y| \text{ units}

Distance of P(x,y) from Y-axis=∣x∣ units\text{Distance of } P(x, y) \text{ from } Y\text{-axis} = |x| \text{ units}

Area of a triangle=12×base×height\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}

Length of a horizontal segment between (x1,y) and (x2,y)=∣x2−x1∣\text{Length of a horizontal segment between } (x_1, y) \text{ and } (x_2, y) = |x_2 - x_1|

Length of a vertical segment between (x,y1) and (x,y2)=∣y2−y1∣\text{Length of a vertical segment between } (x, y_1) \text{ and } (x, y_2) = |y_2 - y_1|

💡Examples

Problem 1:

Find the area of a triangle whose vertices are A(2,0)A(2, 0), B(−2,0)B(-2, 0), and C(0,4)C(0, 4).

Solution:

  1. Identify the base: The points A(2,0)A(2, 0) and B(−2,0)B(-2, 0) lie on the XX-axis. Length of base AB=∣2−(−2)∣=∣2+2∣=4AB = |2 - (-2)| = |2 + 2| = 4 units.
  2. Identify the height: The vertex C(0,4)C(0, 4) lies on the YY-axis. The vertical distance from the origin (which lies on the base ABAB) to point CC is the height. Height h=∣4−0∣=4h = |4 - 0| = 4 units.
  3. Calculate the area: Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} Area=12×4×4\text{Area} = \frac{1}{2} \times 4 \times 4 Area=8 square units\text{Area} = 8 \text{ square units}

Explanation:

Since two vertices lie on the XX-axis, the segment connecting them is the base. The yy-coordinate of the third vertex represents the height relative to the XX-axis.

Problem 2:

Three vertices of a rectangle are P(2,2)P(2, 2), Q(−3,2)Q(-3, 2), and R(−3,−2)R(-3, -2). Find the coordinates of the fourth vertex SS and the area of the rectangle.

Solution:

  1. Analyze coordinates: For a rectangle, opposite sides are parallel and equal.
  • P(2,2)P(2, 2) and Q(−3,2)Q(-3, 2) form a horizontal line (same yy-coordinate).
  • Q(−3,2)Q(-3, 2) and R(−3,−2)R(-3, -2) form a vertical line (same xx-coordinate).
  1. Find SS: The vertex SS must have the same xx-coordinate as PP and the same yy-coordinate as RR to complete the rectangle. Coordinates of S=(2,−2)S = (2, -2).
  2. Calculate dimensions: Length PQ=∣2−(−3)∣=5PQ = |2 - (-3)| = 5 units. Breadth QR=∣2−(−2)∣=4QR = |2 - (-2)| = 4 units.
  3. Calculate area: Area=length×breadth\text{Area} = \text{length} \times \text{breadth} Area=5×4=20 square units\text{Area} = 5 \times 4 = 20 \text{ square units}

Explanation:

In a rectangle, vertices follow a specific symmetry. By aligning the xx and yy coordinates of the given points, the missing vertex and side lengths are determined.

Problem 3:

Determine which quadrant or axis the following points lie in: M(−5,0)M(-5, 0), N(0,3)N(0, 3), O(−2,−2)O(-2, -2), and P(4,−1)P(4, -1).

Solution:

  1. M(−5,0)M(-5, 0): Since the yy-coordinate is 00, the point lies on the negative XX-axis.
  2. N(0,3)N(0, 3): Since the xx-coordinate is 00, the point lies on the positive YY-axis.
  3. O(−2,−2)O(-2, -2): Both xx and yy are negative. This point lies in Quadrant III.
  4. P(4,−1)P(4, -1): xx is positive and yy is negative. This point lies in Quadrant IV.

Explanation:

Points with a zero coordinate lie on the axes, while non-zero coordinates determine the quadrant based on the sign of the abscissa and ordinate.

Problem 4:

A square ABCDABCD has its center at the origin O(0,0)O(0, 0) and its sides are parallel to the axes. If the coordinates of vertex AA are (3,3)(3, 3), find the coordinates of the other three vertices and calculate the perimeter of the square.

Square ABCD centered at the origin with vertices at (3,3), (-3,3), (-3,-3), and (3,-3).

Solution:

  1. Since the sides are parallel to the axes and the center is at (0,0)(0, 0), the square is symmetric across both axes.
  2. Given A(3,3)A(3, 3) is in Quadrant I, the other vertices are found by reflecting across the axes.
  3. BB (reflection of AA across YY-axis) = (−3,3)(-3, 3).
  4. CC (reflection of BB across XX-axis) = (−3,−3)(-3, -3).
  5. DD (reflection of AA across XX-axis) = (3,−3)(3, -3).
  6. The side length ss is the distance between A(3,3)A(3, 3) and B(−3,3)B(-3, 3), which is ∣3−(−3)∣=6|3 - (-3)| = 6 units.
  7. Perimeter=4×s=4×6=24\text{Perimeter} = 4 \times s = 4 \times 6 = 24 units.

Explanation:

In a square centered at the origin with sides parallel to the axes, the vertices will have coordinates (±x,±y)(\pm x, \pm y). Here, x=3x=3 and y=3y=3.

Problem 5:

Find the area of the figure formed by joining the points A(0,5)A(0, 5), B(−4,0)B(-4, 0), and C(4,0)C(4, 0).

Triangle with vertices at (0,5), (-4,0), and (4,0).

Solution:

  1. Plot the points on the Cartesian plane. B(−4,0)B(-4, 0) and C(4,0)C(4, 0) lie on the XX-axis, while A(0,5)A(0, 5) lies on the YY-axis.
  2. The figure ABCABC is a triangle.
  3. The base BCBC lies on the XX-axis. Length of base BC=∣4−(−4)∣=8\text{Length of base } BC = |4 - (-4)| = 8 units.
  4. The height is the perpendicular distance from vertex AA to the base BCBC (the XX-axis). Since AA is at (0,5)(0, 5), the height h=5h = 5 units.
  5. Area=12×base×height=12×8×5=20 sq. units\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 5 = 20 \text{ sq. units}.

Explanation:

Since two vertices lie on the x-axis and the third on the y-axis, the height of the triangle is simply the y-coordinate of the vertex on the y-axis.

The Cartesian System-advanced Class 9 Notes & Examples