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Coordinate Geometry - Introduction: The Language of Graphs-advanced

Grade 9CBSE

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

🔑Concepts

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The Cartesian Plane is divided into four regions called quadrants by the intersection of the horizontal xx-axis and the vertical yy-axis at the origin O(0,0)O(0, 0). In the first quadrant (II), both xx and yy are positive; in the second (IIII), x<0x < 0 and y>0y > 0; in the third (IIIIII), both are negative; and in the fourth (IVIV), x>0x > 0 and y<0y < 0.

Cartesian plane showing four quadrants with signs of coordinates.
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Any point PP on the plane is represented by an ordered pair (x,y)(x, y). The first number xx, the abscissa, denotes the perpendicular distance from the yy-axis. The second number yy, the ordinate, denotes the perpendicular distance from the xx-axis. For any point on the xx-axis, y=0y = 0, and for any point on the yy-axis, x=0x = 0.

A point P showing perpendicular distances to the axes as coordinates.
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Advanced geometry problems often involve calculating areas of shapes formed by connecting plotted points. The area of a rectangle with sides parallel to the axes is the product of the absolute differences of the coordinates: Area=∣x2−x1∣×∣y2−y1∣\text{Area} = |x_2 - x_1| \times |y_2 - y_1|. For triangles, the base and height are determined using the fixed coordinate distances between vertices.

Rectangle plotted on a coordinate plane to demonstrate area calculation.
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Points on a plane can be used to describe reflections. A point P(x,y)P(x, y) reflected across the xx-axis becomes P′(x,−y)P'(x, -y). If reflected across the yy-axis, it becomes P′′(−x,y)P''(-x, y). Reflection through the origin changes the signs of both coordinates to P′′′(−x,−y)P'''(-x, -y).

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Linear relationships can be visualized by plotting solutions of an equation like ax+by+c=0ax + by + c = 0. These points will always lie on a straight line. Horizontal lines are represented by y=ky = k, while vertical lines are represented by x=kx = k.

📐Formulae

P=(x,y)P = (x, y) where xx is the abscissa and yy is the ordinate.

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

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

Area of a Rectangle=∣x2−x1∣×∣y2−y1∣\text{Area of a Rectangle} = |x_2 - x_1| \times |y_2 - y_1|

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

💡Examples

Problem 1:

Plot the points A(4,4)A(4, 4), B(−4,4)B(-4, 4), C(−4,−4)C(-4, -4), and D(4,−4)D(4, -4) on a Cartesian plane. Identify the figure formed and calculate its area.

Solution:

  1. Plotting the points: AA is in Quadrant I, BB is in Quadrant II, CC is in Quadrant III, and DD is in Quadrant IV.
  2. Joining the points A,B,C,DA, B, C, D in order forms a square.
  3. The length of side ABAB is the horizontal distance: ∣4−(−4)∣=8|4 - (-4)| = 8 units.
  4. The length of side BCBC is the vertical distance: ∣4−(−4)∣=8|4 - (-4)| = 8 units.
  5. Since all sides are equal and perpendicular, the figure is a square.
  6. Area calculation: Area=side×sideArea = \text{side} \times \text{side} Area=8×8=64 square unitsArea = 8 \times 8 = 64 \text{ square units}

Explanation:

We use the absolute difference between coordinates to find the lengths of the sides of the geometric figure formed on the grid.

Problem 2:

Determine the coordinates of a point PP that lies in the second quadrant, is 33 units away from the xx-axis and 55 units away from the yy-axis.

Solution:

  1. In the second quadrant, the xx-coordinate is negative and the yy-coordinate is positive.
  2. The distance from the yy-axis corresponds to the absolute value of the xx-coordinate: ∣x∣=5|x| = 5. Since it is in the second quadrant, x=−5x = -5.
  3. The distance from the xx-axis corresponds to the absolute value of the yy-coordinate: ∣y∣=3|y| = 3. Since it is in the second quadrant, y=3y = 3.
  4. Therefore, the coordinates are (−5,3)(-5, 3).

Explanation:

The distance from an axis is the absolute value of the opposite coordinate. Quadrant signs determine if the value is positive or negative.

Problem 3:

Find the area of a triangle whose vertices are O(0,0)O(0,0), A(6,0)A(6,0), and B(0,8)B(0,8).

Solution:

  1. Vertex O(0,0)O(0,0) is the origin.
  2. Vertex A(6,0)A(6,0) lies on the xx-axis. The length of the base OAOA is ∣6−0∣=6|6 - 0| = 6 units.
  3. Vertex B(0,8)B(0,8) lies on the yy-axis. The length of the height OBOB is ∣8−0∣=8|8 - 0| = 8 units.
  4. The triangle OABOAB is a right-angled triangle because the xx and yy axes are perpendicular.
  5. Area calculation: Area=12×base×heightArea = \frac{1}{2} \times \text{base} \times \text{height} Area=12×6×8Area = \frac{1}{2} \times 6 \times 8 Area=24 square unitsArea = 24 \text{ square units}

Explanation:

When vertices lie on the axes, the lengths from the origin act as the base and height for the area formula.

Problem 4:

Points A(−2,0)A(-2, 0), B(2,0)B(2, 0), and C(0,23)C(0, 2\sqrt{3}) are vertices of a triangle. Show that the triangle is equilateral and find its area.

An equilateral triangle ABC plotted with base on the x-axis.

Solution:

  1. Plot the points: A(−2,0)A(-2, 0) and B(2,0)B(2, 0) lie on the xx-axis. C(0,23)C(0, 2\sqrt{3}) lies on the yy-axis.
  2. Calculate side lengths: AB=∣2−(−2)∣=4AB = |2 - (-2)| = 4 units. Using Pythagoras on △OBC\triangle OBC (where OO is origin): BC=22+(23)2=4+12=16=4BC = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4. Similarly, AC=(−2)2+(23)2=4AC = \sqrt{(-2)^2 + (2\sqrt{3})^2} = 4.
  3. Since AB=BC=AC=4AB = BC = AC = 4, the triangle is equilateral.
  4. Area = 12×base×height=12×4×23=43\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2\sqrt{3} = 4\sqrt{3} sq units.

Explanation:

We use the coordinates to determine side lengths. Since the base lies on the xx-axis, the yy-coordinate of the third vertex serves as the altitude.

Problem 5:

A square has two of its vertices at (1,2)(1, 2) and (4,2)(4, 2). If the square lies entirely in the first quadrant, find the coordinates of the other two vertices and the coordinates of the center of the square.

A square ABCD in the first quadrant with side length 3.

Solution:

  1. The side length ss is the distance between (1,2)(1, 2) and (4,2)(4, 2): s=∣4−1∣=3s = |4 - 1| = 3.
  2. Since it is a square and lies in the first quadrant, we move 33 units vertically up from both points.
  3. Vertex D=(1,2+3)=(1,5)D = (1, 2 + 3) = (1, 5).
  4. Vertex C=(4,2+3)=(4,5)C = (4, 2 + 3) = (4, 5).
  5. The center MM is the midpoint of diagonal ACAC: x=1+42=2.5x = \frac{1+4}{2} = 2.5, y=2+52=3.5y = \frac{2+5}{2} = 3.5.
  6. Center M=(2.5,3.5)M = (2.5, 3.5).

Explanation:

Since the segment joining (1,2)(1, 2) and (4,2)(4, 2) is horizontal, the vertical sides must have the same length. We add the side length to the yy-coordinates to stay in the first quadrant.