Madhyamik Project on Coordinate Geometry | Class 9

Introduction

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that combines elements of both algebra and geometry. It provides a way to study geometric shapes and their properties using algebraic methods and coordinates. In coordinate geometry, points in a plane are represented by ordered pairs of numbers, typically (x, y), where “x” represents the horizontal position (along the x-axis), and “y” represents the vertical position (along the y-axis).

Axis of Co-ordinate Axis

In the context of coordinate geometry, the term “axis” typically refers to the two perpendicular number lines that define the Cartesian coordinate system. These two lines are known as the “coordinate axes,” and they are used to specify the position of points in a plane. There are two coordinate axes:

1. X-Axis: The horizontal number line is called the x-axis. It extends infinitely in both the positive (+x) and negative (-x) directions from the origin, which is the point where the x-axis and y-axis intersect. The x-axis is used to represent the horizontal position of a point.
2. Y-Axis: The vertical number line is called the y-axis. Similar to the x-axis, it extends infinitely in both the positive (+y) and negative (-y) directions from the origin. The y-axis is used to represent the vertical position of a point.

Different Quadrants with their Sign

In the Cartesian coordinate system, the plane is divided into four quadrants, and each quadrant has specific characteristics regarding the signs of the coordinates (x, y). Here are the four quadrants and their associated signs:

Quadrant I (QI):

• The first quadrant is located in the upper right-hand part of the Cartesian plane.
• Both the x-coordinate (horizontal) and y-coordinate (vertical) are positive in this quadrant.
• Points in this quadrant have coordinates (x, y) where both x > 0 and y > 0.

Quadrant II (QII):

• The second quadrant is located in the upper left-hand part of the Cartesian plane.
• The x-coordinate is negative (x < 0), but the y-coordinate is positive (y > 0) in this quadrant.
• Points in this quadrant have coordinates (x, y) where x < 0 and y > 0.

Quadrant III (QIII):

• The third quadrant is located in the lower left-hand part of the Cartesian plane.
• Both the x-coordinate and y-coordinate are negative in this quadrant.
• Points in this quadrant have coordinates (x, y) where both x < 0 and y < 0.

Quadrant IV (QIV):

• The fourth quadrant is located in the lower right-hand part of the Cartesian plane.
• The x-coordinate is positive (x > 0), but the y-coordinate is negative (y < 0) in this quadrant.
• Points in this quadrant have coordinates (x, y) where x > 0 and y < 0.

Distance Formula

Let P(x₁, y₁) and Q(x₂, y₂) be the coordinates of two points on the coordinate plane.

Draw two lines parallel to both the x-axis and y-axis (as shown in the figure) through P and Q.

The parallel line through P will meet the perpendicular drawn to the x-axis from Q at T.

Thus, ΔPTQ  is right-angled at T.

PT = Base, QT = Perpendicular and PQ = Hypotenuse

By Pythagoras’s Theorem,

PQ² = PT² + QT²

= (x₂ – x₁)² + (y₂ – y₁)²

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²]

Hence, the distance between two points (x₁, y₁) and (x₂, y₂) is √[(x₂ – x₁)² + (y₂ – y₁)²]

Similarly, the distance of a point P(x, y) from the origin O(0, 0) in the Cartesian plane is given by the formula:

OP = √(x² + y²)

Example Question

In a classroom, four friends are seated at points A, B, C and D. Anku and Ankila walk into the classroom and after observing for some time Anku asks Ankita. Don’t you think ABCD is a square? Ankita disagrees. Who is correct?

Answer

According to fig. A(3,4),B(6,7),C(9,4) and D(6,1) are the position of 4 friends .

If we join all these points then

By using the distance formula

= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AB = \sqrt{(3-6)^2+(4-7)^2} = 3√2

BC = \sqrt{(6-9)^2+(7-4)^2} = 3√2

CB = \sqrt{(9-6)^2+(4-1)^2} = 3√2

AD = \sqrt{(3-6)^2+(4-1)^2} = 3√2

If we join the point $AC$ and then

Diagonal AC = \sqrt{(3-9)^2+(4-4)^2} = √36 = 6

Diagonal AC = \sqrt{(6-6)^2+(7-1)^2} = √36 = 6

Hence Anku is correct.

Plot the following points and write the name of the figure obtained by joining, them in order P(-3, 2), Q(-7, -3), R(6, -3) and S(2, 2). Also, find the area of the figure.

Answer

The shape of the figure is the Trapezium

Area of Trapezium

= ½ × sum of parallel sides × height

= ½ × (5 + 13) × 5

= ½ ×18 × 5 = 45 sq unit.

The following observed values of x and y are thought to satisfy a linear equation. Write the line equation.

Draw the graph using the values of x, and y given in the above table. At what points does the graph of the linear equation (i) cut the x-axis and (ii) cut the y-axis?

Answer

The linear equation is 2x+3y=6. Both the points (6,−2) and (−6,6) satisfy the given linear equation.

Plot the points (6,−2) and (−6,6) on a graph paper. Now join these two points and obtain a line. We see that the graph cuts the x-axis at (3,0) and the y-axis at (0,2).

Acknowledgement

I would like to express my special gratitude to my English teacher _____ (teacher’s name) as well as our principal _____ (principal’s name) who gave me the golden opportunity to do this wonderful project on _____ (write topic Name) Secondly, I would also like to thank my parents and friends who helped me a lot in finalizing this project within the limited time frame.

Lastly, I would like to thank all my supporters who motivated me to complete my project before the deadline.

Bibliography

• Flash Education – A useful website with educational content. (Website: FlashEducation.online)
• Wikipedia – An online encyclopedia where I found useful facts. (Website: Wikipedia)
• Understanding Mathematics – A mathematics book of class 9.