CBSE Class 10 Maths Chapter 7 Coordinate Geometry Notes

Key Formulas and Concepts to Remember

• Distance Formula: The distance between two points P(x₁, y₁) and Q(x₂, y₂) is given by PQ = √((x₂ - x₁)² + (y₂ - y₁)²).
• Distance from Origin: The distance of a point P(x, y) from the origin O(0, 0) is √(x² + y²).
• Collinear Points: Three points A, B, and C are collinear if the sum of the distances between two pairs of points is equal to the distance of the third pair (e.g., AB + BC = AC). Alternatively, if the area of the triangle formed by them is zero.
• Section Formula (Internal Division): The coordinates of a point P(x, y) that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁:m₂ are P(x, y) = ((m₁x₂ + m₂x₁) / (m₁ + m₂), (m₁y₂ + m₂y₁) / (m₁ + m₂)).
• Midpoint Formula: A special case of the section formula where m₁ = m₂ = 1. The midpoint M(x, y) of the line segment joining A(x₁, y₁) and B(x₂, y₂) is M(x, y) = ((x₁ + x₂) / 2, (y₁ + y₂) / 2).
• Area of a Triangle: The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. The absolute value ensures a positive area.
• Centroid of a Triangle: The coordinates of the centroid of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) are ((x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3). (This is generally taught as an extension, good for competitive exams or higher-level understanding).

Key Terms in Coordinate Geometry

Coordinate Plane (Cartesian Plane)

A two-dimensional plane defined by two perpendicular number lines, the x-axis (horizontal) and the y-axis (vertical), intersecting at the origin.

Origin

The point where the x-axis and y-axis intersect, represented by coordinates (0, 0).

Coordinates

An ordered pair (x, y) that specifies the location of a point in the coordinate plane. 'x' is the abscissa and 'y' is the ordinate.

Abscissa

The x-coordinate of a point, representing its horizontal distance from the y-axis.

Ordinate

The y-coordinate of a point, representing its vertical distance from the x-axis.

Quadrants

The four regions into which the coordinate plane is divided by the x-axis and y-axis. They are numbered I, II, III, and IV in a counter-clockwise direction.

Collinear Points

Three or more points that lie on the same straight line.

Understanding the Distance Formula

The Distance Formula is a cornerstone of coordinate geometry, allowing us to calculate the straight-line distance between any two points in a plane. Its derivation is rooted in the Pythagorean Theorem. Imagine two points, P(x₁, y₁) and Q(x₂, y₂). We can form a right-angled triangle by drawing a horizontal line from P and a vertical line from Q, meeting at a third point R(x₂, y₁).

The horizontal distance PR is simply the difference in their x-coordinates, |x₂ - x₁|. Similarly, the vertical distance QR is the difference in their y-coordinates, |y₂ - y₁|. According to the Pythagorean Theorem, the square of the hypotenuse (PQ) is equal to the sum of the squares of the other two sides (PR and QR). So, PQ² = PR² + QR².

Substituting the distances, we get PQ² = (x₂ - x₁)² + (y₂ - y₁)². Taking the square root of both sides gives us the Distance Formula: PQ = √((x₂ - x₁)² + (y₂ - y₁)²). It's important to remember that the order of subtraction (x₁ - x₂ vs x₂ - x₁) doesn't matter because the difference is squared, resulting in a positive value. This formula is widely used to check if a given figure is a square, rectangle, rhombus, or parallelogram, by comparing side lengths and diagonals, and also to prove collinearity of points.

Steps to Apply the Section Formula

1. Identify the Endpoints and Ratio — Clearly identify the coordinates of the two endpoints of the line segment, say A(x₁, y₁) and B(x₂, y₂). Also, identify the ratio m₁:m₂ in which the point P divides the segment.
2. Assign Variables to Coordinates — Label x₁, y₁, x₂, y₂, m₁, and m₂ with their respective values from the problem statement. Double-check for correct assignment.
3. Apply the Formula for X-coordinate — Use the formula for the x-coordinate of the dividing point P: x = (m₁x₂ + m₂x₁) / (m₁ + m₂). Substitute the values and calculate carefully.
4. Apply the Formula for Y-coordinate — Use the formula for the y-coordinate of the dividing point P: y = (m₁y₂ + m₂y₁) / (m₁ + m₂). Substitute the values and calculate.
5. State the Final Coordinates — Combine the calculated x and y values to state the coordinates of the point P(x, y). Remember to always write coordinates as an ordered pair.
6. Special Case: Midpoint — If the point is the midpoint, the ratio is 1:1 (m₁ = m₂ = 1). The formula simplifies to ((x₁ + x₂) / 2, (y₁ + y₂) / 2).

Worked Examples

• {"title":"Example 1: Distance Calculation","description":"Find the distance between the points A(2, 3) and B(4, 1).\n\nSolution:\nGiven (x₁, y₁) = (2, 3) and (x₂, y₂) = (4, 1).\nUsing the distance formula, AB = √((4 - 2)² + (1 - 3)²)\nAB = √((2)² + (-2)²)\nAB = √(4 + 4)\nAB = √8 = 2√2 units."}
• {"title":"Example 2: Section Formula Application","description":"Find the coordinates of the point which divides the line segment joining (4, -3) and (8, 5) in the ratio 3:1 internally.\n\nSolution:\nLet the points be A(4, -3) and B(8, 5). Ratio m₁:m₂ = 3:1.\nUsing the section formula P(x, y) = ((m₁x₂ + m₂x₁) / (m₁ + m₂), (m₁y₂ + m₂y₁) / (m₁ + m₂))\nx = (38 + 14) / (3 + 1) = (24 + 4) / 4 = 28 / 4 = 7\ny = (35 + 1(-3)) / (3 + 1) = (15 - 3) / 4 = 12 / 4 = 3\nSo, the coordinates of the point are (7, 3)."}
• {"title":"Example 3: Area of a Triangle","description":"Find the area of the triangle with vertices (1, -1), (-4, 6), and (-3, -5).\n\nSolution:\nGiven (x₁, y₁) = (1, -1), (x₂, y₂) = (-4, 6), (x₃, y₃) = (-3, -5).\nArea = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|\nArea = ½ |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|\nArea = ½ |1(11) - 4(-4) - 3(-7)|\nArea = ½ |11 + 16 + 21|\nArea = ½ |48|\nArea = 24 square units."}

Exam Tips for Coordinate Geometry

1. Memorize Formulas: Write down the Distance, Section, Midpoint, and Area of Triangle formulas repeatedly. Knowing them perfectly saves time and prevents errors.
2. Careful with Signs: A common mistake is getting signs wrong (especially with negative coordinates). Pay close attention when substituting values into formulas.
3. Draw a Diagram: For complex problems (e.g., proving a type of quadrilateral, finding a vertex), sketching a rough diagram can help visualize the problem and identify relationships between points.
4. Collinearity Check: Remember two ways to check collinearity: either the sum of two distances equals the third, or the area of the triangle formed by the three points is zero. The latter is often quicker.
5. Section Formula (Ratio): If a point divides a line segment into k:1, you can use k:1 instead of m₁:m₂ for easier calculation. This is particularly useful when you need to find the ratio itself.
6. Read Questions Carefully: Note keywords like 'midpoint', 'ratio', 'on the x-axis' (y=0), 'on the y-axis' (x=0). These provide crucial information.

Quick Revision Checks

• Q: What is the distance of the point (-5, 12) from the origin? A: Using the distance formula from the origin √(x² + y²), the distance is √((-5)² + 12²) = √(25 + 144) = √169 = 13 units.
• Q: If a point P divides the line segment joining A(1, 7) and B(4, -3) in the ratio 2:3 internally, what are the coordinates of P? A: Using the section formula, x = (24 + 31)/(2+3) = 11/5; y = (2(-3) + 37)/(2+3) = 15/5 = 3. So, P(11/5, 3).
• Q: What is the midpoint of the line segment joining (3, 4) and (-1, 2)? A: Using the midpoint formula, x = (3 + (-1))/2 = 2/2 = 1; y = (4 + 2)/2 = 6/2 = 3. The midpoint is (1, 3).
• Q: How do you determine if three given points are collinear? A: Three points are collinear if the area of the triangle formed by them is zero. Alternatively, if the sum of the distances between two pairs of points equals the distance of the third pair (e.g., AB + BC = AC).

Frequently Asked Questions