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Class 10 · Maths · Chapter 7 · NCERT Class 10 Mathematics

Coordinate Geometry Class 10 Notes

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Chapter mind map

The whole chapter at a glance: the big idea, then each branch and what sits under it.

Coordinate Geometry

A bridge between algebra and geometry using a Cartesian plane to represent shapes and calculate spatial relationships.

  1. Distance Formula

    Calculates the length of a line segment between two points using horizontal and vertical displacements.

    • Pythagorean Derivation — Derived from PQ² = (x₂ - x₁)² + (y₂ - y₁)² where sides represent absolute differences in coordinates.
    • Distance from Origin — Simplified formula OP = √(x² + y²) when one point is (0, 0).
    • Geometric Derivation Steps — Identify points, construct right triangle PTQ, calculate side lengths, and apply Pythagoras Theorem.

  2. Geometric Verifications

    Using side lengths and diagonals to identify specific geometric properties and shapes.

    • Collinear Points — Three points are collinear if the sum of two distances equals the third (AB + BC = AC).
    • Triangle Properties — Sum of any two sides must be > third side. Right triangles satisfy a² + b² = c².
    • Quadrilateral Types — Squares have equal sides and diagonals; Rhombuses have equal sides but unequal diagonals.
    • Rhombus Area — Calculated as (d₁ × d₂)/2 using distances of diagonals.

  3. Section Formula

    Determines coordinates of a point P dividing a segment AB in a specific ratio m₁ : m₂.

    • Internal Division — x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂).
    • Mid-point Formula — Special case where ratio is 1:1; coordinates are averages of endpoints: ((x₁+x₂)/2, (y₁+y₂)/2).
    • Parallelogram Property — Diagonals bisect each other; mid-points of both diagonals are identical.

  4. Ratio and Trisection

    Methods for finding unknown ratios or multiple division points along a segment.

    • K : 1 Method — Simplifies ratio calculations by using a single variable k; if k = 2/7, the ratio is 2:7.
    • Trisection Points — Dividing a segment into three equal parts using ratios 1:2 and 2:1.
    • Multiple Equal Parts — For four parts, find the main mid-point, then mid-points of the resulting halves.

  5. Equidistant Relations

    Solving for points or algebraic relationships where distances to two fixed points are equal.

    • Perpendicular Bisector — The set of all points equidistant from A and B forms this line.
    • Algebraic Method — Set AP² = BP² to remove square roots and solve for the relationship between x and y.
    • Axis-Specific Points — Points on x-axis are (x, 0); points on y-axis are (0, y).

Chapter notes

An exploration of algebraic tools used to study geometric figures, focusing on the distance between points and the division of line segments in specific ratios.

01 · Explore

Introduction to Coordinate Geometry

Coordinate geometry serves as a bridge between algebra and geometry, allowing us to represent geometric shapes using algebraic equations.

In earlier studies, we learned that the position of a point in a plane is determined by its distance from two perpendicular axes: the x-axis and the y-axis. The distance of a point from the y-axis is its x-coordinate (abscissa), and its distance from the x-axis is its y-coordinate (ordinate). Together, these form the coordinates (x, y).

Points on the x-axis always have a y-coordinate of 0, represented as (x, 0). Conversely, points on the y-axis have an x-coordinate of 0, represented as (0, y). This system allows us to visualize linear equations as straight lines and quadratic equations as parabolas.

Coordinate geometry is not just theoretical; it is an essential tool in physics, engineering, navigation, and even seismology. In this chapter, we focus on two primary objectives: calculating the distance between two points and finding the coordinates of a point that divides a line segment in a given ratio.

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Question

What are the coordinates of a point that lies on the y-axis at a distance of 5 units from the x-axis in the positive direction?

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NCERT reference: chapter PDF page 1.

02 · Explore

Deriving the Distance Formula

The distance formula is derived using the Pythagoras Theorem to find the length of the line segment connecting any two points in a Cartesian plane.

To find the distance between two points P(x₁, y₁) and Q(x₂, y₂), we construct a right-angled triangle. Draw perpendiculars PR and QS to the x-axis. Then, draw a perpendicular PT from P to the line QS. This creates a right triangle PTQ.

The horizontal length PT is |x₂ − x₁| and the vertical length QT is |y₂ − y₁|. The bars mean absolute value: side lengths cannot be negative. Squaring either signed difference gives the same positive result. By Pythagoras, PQ² = PT² + QT².

Therefore, PQ² = PT² + QT², which simplifies to PQ² = (x₂ - x₁)² + (y₂ - y₁)². Taking the positive square root (since distance cannot be negative), we get the Distance Formula: PQ = √((x₂ - x₁)² + (y₂ - y₁)²).

A(1, 1)B(5, 4)4 units35 unitsxy
From A(1, 1) to B(5, 4), move 4 units horizontally and 3 vertically. The distance is √(4² + 3²) = 5 units. Learning sketch; use the labels and stated dimensions, not measurements from the picture.

Steps to Derive Distance PQ

  1. 1

    Identify Points

    Locate P(x1, y1) and Q(x2, y2) on the plane.

  2. 2

    Construct Triangle

    Draw horizontal and vertical lines to form a right triangle PTQ.

  3. 3

    Calculate Sides

    Horizontal length = |x₂ − x₁|; vertical length = |y₂ − y₁|. Lengths are non-negative.

  4. 4

    Apply Pythagoras

    PQ = √[(x2 - x1)² + (y2 - y1)²].

The geometric derivation of the distance formula using horizontal and vertical displacements.

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Question

Does the order of subtraction matter in the distance formula (e.g., x1 - x2 instead of x2 - x1)?

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NCERT reference: chapter PDF pages 2, 3, 4.

03 · Explore

Distance from the Origin

A special case of the distance formula occurs when one of the points is the origin (0, 0).

When we want to find the distance of a point P(x, y) from the origin O(0, 0), we substitute x₁ = 0, y₁ = 0, x₂ = x, and y₂ = y into the distance formula.

The formula simplifies significantly: OP = √((x - 0)² + (y - 0)²), which results in OP = √(x² + y²). This is a direct application of the Pythagoras Theorem where the coordinates x and y represent the base and height of the triangle. If a coordinate is negative, its absolute value gives the corresponding length; squaring the coordinate already handles the sign.

Calculating Distance from Origin

OP = √(36² + 15²)

To find the distance of point (36, 15) from the origin: Square 36 (1296) and 15 (225). Sum them to get 1521. The square root of 1521 is 39. Thus, the distance is 39 units.

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Question

Find the distance of the point (-6, 8) from the origin.

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NCERT reference: chapter PDF page 4.

04 · Explore

Verifying Geometric Shapes

The distance formula is a powerful tool for identifying types of triangles and quadrilaterals based on their side lengths and diagonals.

To check if three points form a triangle, we use the Triangle Inequality Property: the sum of any two sides must be greater than the third side. If the sum of the squares of two sides equals the square of the third side, the triangle is a right-angled triangle.

For quadrilaterals, we can distinguish between shapes by comparing all four sides and the two diagonals. For example, in a square, all four sides are equal and both diagonals are equal. In a rhombus, all four sides are equal, but the diagonals are usually not equal.

If we find that the sum of two distances (AB + BC) is exactly equal to the third distance (AC), then the points A, B, and C are collinear, meaning they lie on the same straight line.

For a rhombus, you can also use the distance formula to find the diagonals d₁ and d₂, then calculate area = d₁d₂/2. Keep the order of the vertices consistent so that you identify diagonals rather than sides.

Proving a Square

AB = BC = CD = DA = √34; AC = BD = √68

Given points A(1, 7), B(4, 2), C(-1, -1), and D(-4, 4). Calculating all side lengths using the distance formula gives √34 for each. Calculating diagonals AC and BD gives √68 for each. Since all sides are equal and diagonals are equal, ABCD is a square.

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Question

If AB = 3, BC = 4, and AC = 7, are the points A, B, and C collinear?

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NCERT reference: chapter PDF pages 4, 5, 6, 13.

05 · Explore

Equidistant Points and Relations

We can find specific points or algebraic relationships by setting the distances between a variable point and two fixed points as equal.

If a point P(x, y) is equidistant from two points A and B, then AP = BP. Squaring both sides (AP² = BP²) allows us to remove the square root and solve for a relationship between x and y.

The set of all points equidistant from A and B forms the perpendicular bisector of the line segment AB. This means any point (x, y) satisfying the resulting linear equation lies on this bisector.

When asked to find a point on a specific axis that is equidistant from two points, remember to use the property of that axis. For the x-axis, the point is (x, 0); for the y-axis, the point is (0, y).

Finding a Point on the Y-axis

(6 - 0)² + (5 - y)² = (-4 - 0)² + (3 - y)²

To find a point P(0, y) equidistant from A(6, 5) and B(-4, 3), set AP² = BP². Expanding gives 36 + 25 - 10y + y² = 16 + 9 - 6y + y². Simplifying: 61 - 10y = 25 - 6y. Solving for y: 4y = 36, so y = 9. The point is (0, 9).

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What is the first step to find a point on the x-axis equidistant from two given points?

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NCERT reference: chapter PDF pages 6, 7.

The rest of this chapter

Keep reading Coordinate Geometry, free

  1. Locked: 1. The Section Formula
  2. Locked: 2. The Mid-point Formula
  3. Locked: 3. Finding the Ratio
  4. Locked: 4. Trisection and Multiple Divisions

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