Coordinate Geometry Class 10 Notes PDF: Chapter 7 Formulas, Examples and Questions

These Coordinate Geometry Class 10 Notes PDF cover the current CBSE and NCERT Chapter 7 syllabus. The main topics are the coordinate plane, distance formula, section formula for internal division and midpoint formula.

This guide also includes solved examples, common mistakes, Basic and Standard practice questions, a formula-selection guide and clear information about deleted or higher-class topics. Reviewed using the CBSE Class 10 Mathematics curriculum for 2026–27 and the current NCERT Coordinate Geometry chapter.

Coordinate Geometry Class 10 Syllabus for 2026–27

The official CBSE Class 10 syllabus includes a review of coordinate geometry, the distance formula and the section formula for internal division.

Topics Included in the Current Syllabus

Students should be able to:

1. Read and plot points on a coordinate plane.
2. Identify the x-axis, y-axis, origin and quadrants.
3. Find the distance between two points.
4. Find the distance of a point from the origin.
5. Find a point that divides a line segment internally in a given ratio.
6. Use the midpoint formula as the 1:1 case of internal division.
7. Apply these formulas to geometrical and real-life questions.

The Coordinate Geometry unit is assigned six marks in the CBSE Class 10 course structure for 2026–27. This does not guarantee one fixed six-mark question because the number and type of questions may vary.

Download Coordinate Geometry Class 10 Notes PDF

The downloadable Coordinate Geometry notes should contain complete explanations, formulas, diagrams, solved examples, practice questions and answers without requiring registration.

Current, Derived and Deleted Topics

• Coordinate-plane review: Current syllabus. Learn fully.
• Distance formula: Current syllabus. Learn fully.
• Internal section formula: Current syllabus. Learn fully.
• Midpoint formula: Special case of internal division. Learn fully.
• Equidistant-point problems: Application of the distance formula. Practise.
• Trisection and equal division: Application of the section formula. Practise.
• Area of a triangle using coordinates: Rationalised out of the current chapter.
• External division: Higher-class topic. Do not mix it with the Class 10 internal formula.
• Centroid: Enrichment or higher-level topic.
• Slope and equation of a line: Outside the present Chapter 7 scope.

NCERT’s rationalised-content list identifies the coordinate-based “Area of a Triangle” section as dropped content.

The current NCERT chapter teaches internal division and states that external division will be studied in higher classes.

Also Check: CBSE Class 10 Maths Syllabus | CBSE Class 8 Maths Notes 

Coordinate Geometry Basics

Coordinate geometry uses ordered pairs and algebra to locate points and study geometrical relationships on a plane.

Cartesian Plane, Axes and Origin

A Cartesian plane is formed by two perpendicular number lines.

• x-axis: The horizontal number line.
• y-axis: The vertical number line.
• Origin: The point where the two axes meet, written as O(0, 0).

The axes divide the plane into four regions called quadrants.

Ordered Pairs, Abscissa and Ordinate

A point is written as an ordered pair: P(x, y).

• Abscissa: The x-coordinate, which gives the horizontal position.
• Ordinate: The y-coordinate, which gives the vertical position.

The order matters. The points (2, 5) and (5, 2) are in different positions.

Signs in the Four Quadrants

• Quadrant I: (+, +), for example (3, 4)
• Quadrant II: (−, +), for example (−3, 4)
• Quadrant III: (−, −), for example (−3, −4)
• Quadrant IV: (+, −), for example (3, −4)

A useful order to remember is:

(+, +), (−, +), (−, −), (+, −)

Coordinates of Points on the Axes

• Point on the x-axis: (x, 0)
• Point on the y-axis: (0, y)
• Origin: (0, 0)

Common mistake: A point on the x-axis does not have an x-coordinate of zero. Its y-coordinate is zero because its vertical distance from the x-axis is zero.

Distance from the x-Axis and y-Axis

For a point P(x, y):

• Distance from the x-axis: |y|
• Distance from the y-axis: |x|

Absolute values are used because distance cannot be negative.

Example: For P(−6, 4):

• Distance from the x-axis = |4| = 4 units
• Distance from the y-axis = |−6| = 6 units

Reflection of a Point

• Reflection across the x-axis: (x, y) becomes (x, −y)
• Reflection across the y-axis: (x, y) becomes (−x, y)
• Reflection across both axes: (x, y) becomes (−x, −y)

Example: The reflection of (−3, 5) across the x-axis is (−3, −5).

Distance Formula Class 10

The distance formula finds the straight-line distance between two points by using their horizontal and vertical coordinate differences.

Distance Between Two Points

For points P(x₁, y₁) and Q(x₂, y₂), the distance is:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

Why the Distance Formula Works

The horizontal and vertical differences between two points form the perpendicular sides of a right-angled triangle.

Horizontal change = x₂ − x₁

Vertical change = y₂ − y₁

Using Pythagoras’ theorem:

PQ² = (x₂ − x₁)² + (y₂ − y₁)²

Therefore:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

Steps for Using the Distance Formula

1. Label the points as (x₁, y₁) and (x₂, y₂).
2. Subtract the x-coordinates.
3. Subtract the y-coordinates.
4. Square both differences.
5. Add the squared values.
6. Take the positive square root.

Worked Example: Points in Different Quadrants

Find the distance between A(−2, 3) and B(4, −5).

AB = √[(4 − (−2))² + (−5 − 3)²]

AB = √[6² + (−8)²]

AB = √(36 + 64)

AB = 10 units

The horizontal difference is 6 and the vertical difference is 8, so a distance of 10 units is reasonable.

Does Subtraction Order Matter?

The subtraction order does not change the final distance because each difference is squared.

(x₂ − x₁)² = (x₁ − x₂)²

For example:

(4 − (−2))² = 6² = 36

(−2 − 4)² = (−6)² = 36

Use the same order for both coordinate pairs to reduce mistakes.

Distance Between Points on the Same Axis

For A(2, 0) and B(9, 0):

AB = |9 − 2| = 7 units

For C(0, −4) and D(0, 3):

CD = |3 − (−4)| = 7 units

The absolute-difference method is faster when both points lie on the same axis.

Distance of a Point from the Origin

For P(x, y), the origin is O(0, 0).

OP = √(x² + y²)

Example: Find the distance of P(−5, 12) from the origin.

OP = √[(-5)² + 12²]

OP = √(25 + 144)

OP = 13 units

Finding an Unknown Coordinate

Find the possible values of y if the distance between P(2, −3) and Q(10, y) is 10 units.

√[(10 − 2)² + (y − (−3))²] = 10

√[64 + (y + 3)²] = 10

64 + (y + 3)² = 100

(y + 3)² = 36

y + 3 = 6 or y + 3 = −6

y = 3 or y = −9

Two answers are possible because two different points can be the same distance from a fixed point.

Point on an Axis Equidistant from Two Points

Find the point on the y-axis that is equidistant from A(2, 3) and B(−4, 1).

A point on the y-axis has the form P(0, y).

Because PA = PB, equate the squared distances:

(0 − 2)² + (y − 3)² = (0 − (−4))² + (y − 1)²

4 + (y − 3)² = 16 + (y − 1)²

4 + y² − 6y + 9 = 16 + y² − 2y + 1

13 − 6y = 17 − 2y

−4 = 4y

y = −1

P = (0, −1)

Why an Equidistant Point Is Not Always the Midpoint

A midpoint is equidistant from the two endpoints, but every equidistant point is not a midpoint.

All points on the perpendicular bisector of a segment are equidistant from its endpoints. The midpoint is only the point where the perpendicular bisector meets the segment.

Use the midpoint formula only when the question says that the point:

• is the midpoint;
• bisects the segment;
• divides the segment in the ratio 1:1; or
• lies halfway between the endpoints.

Applications of the Distance Formula

The distance formula can be used to:

1. Classify a triangle.
2. Check whether a triangle is right-angled.
3. Compare the sides and diagonals of a quadrilateral.
4. Check whether three points are collinear.
5. Find the radius of a circle.

Example: Classify a Triangle

Let A(0, 0), B(3, 4) and C(6, 0).

AB = √(3² + 4²) = 5

BC = √(3² + (−4)²) = 5

AC = 6

Since AB = BC, triangle ABC is an isosceles triangle.

Section Formula Class 10

The section formula gives the coordinates of a point that divides a line segment internally in a given ratio.

Meaning of Internal Division

Suppose P lies between A and B and:

AP : PB = m : n

Then P divides the line segment AB internally in the ratio m:n.

Internal Section Formula

For A(x₁, y₁), B(x₂, y₂) and AP:PB = m:n:

P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

Steps for Applying the Section Formula

1. Write the endpoints in the same order as the diagram.
2. Write the ratio as AP:PB = m:n.
3. Multiply m by the coordinates of B.
4. Multiply n by the coordinates of A.
5. Add the products.
6. Divide both coordinates by m + n.
7. Check that the point lies between the endpoints.

Why the Ratio Multiplies the Opposite Endpoint

The formula uses opposite weighting because a larger first segment places the dividing point closer to the second endpoint.

For example, if AP:PB = 4:1, point P is closer to B. The larger number 4 therefore multiplies the coordinates of B.

Worked Example: Internal Division

Find the point that divides the segment joining A(−2, 4) and B(8, −1) in the ratio 2:3.

m = 2 and n = 3

P = ((2 × 8 + 3 × (−2))/(2 + 3), (2 × (−1) + 3 × 4)/(2 + 3))

P = ((16 − 6)/5, (−2 + 12)/5)

P = (10/5, 10/5)

P = (2, 2)

The m:n Form and the k:1 Form

• m:n: Use m and n directly.
• k:1: Use m = k and n = 1.
• 1:k: Use m = 1 and n = k.

Do not treat k:1 and 1:k as the same ratio.

Reversing the Endpoints

If the endpoint order is reversed, the ratio order must also be reversed.

Original form:

AP:PB = 2:3

Reversed form:

BP:PA = 3:2

Reversing both the endpoints and ratio gives the same dividing point.

Midpoint Formula

A midpoint divides a line segment internally in the ratio 1:1.

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Midpoint Example

Find the midpoint of A(−3, 5) and B(7, −1).

M = ((−3 + 7)/2, (5 + (−1))/2)

M = (4/2, 4/2)

M = (2, 2)

Finding the Ratio When the Point Is Given

The point P(5, 4) divides the segment joining A(1, 2) and B(11, 7). Find AP:PB.

Let AP:PB = m:n.

Using the x-coordinate:

5 = (11m + n)/(m + n)

5m + 5n = 11m + n

4n = 6m

m:n = 2:3

Check using the y-coordinate:

(2 × 7 + 3 × 2)/(2 + 3) = 20/5 = 4

Therefore, AP:PB = 2:3.

Finding an Unknown Endpoint

The midpoint of A(−2, 4) and B(x, y) is M(3, −1). Find B.

For the x-coordinate:

(−2 + x)/2 = 3

−2 + x = 6

x = 8

For the y-coordinate:

(4 + y)/2 = −1

4 + y = −2

y = −6

B = (8, −6)

Finding Where a Segment Meets an Axis

Find the point where the segment joining A(−3, 6) and B(6, −3) meets the x-axis.

Let the point divide AB in the ratio m:n.

Because the point lies on the x-axis, its y-coordinate is zero.

0 = (m × (−3) + n × 6)/(m + n)

−3m + 6n = 0

m = 2n

m:n = 2:1

Now calculate the x-coordinate:

x = (2 × 6 + 1 × (−3))/(2 + 1)

x = 9/3

P = (3, 0)

Points of Trisection

Points of trisection divide a line segment into three equal parts.

For A(0, 0) and B(9, 6):

• The first trisection point divides AB in the ratio 1:2.
• The second trisection point divides AB in the ratio 2:1.

First point:

P = ((1 × 9 + 2 × 0)/3, (1 × 6 + 2 × 0)/3)

P = (3, 2)

Second point:

Q = ((2 × 9 + 1 × 0)/3, (2 × 6 + 1 × 0)/3)

Q = (6, 4)

Applications of the Midpoint Formula

The midpoint formula can be used to:

1. Find the centre of a circle when the endpoints of a diameter are known.
2. Find a missing endpoint when the centre and one endpoint are known.
3. Check whether the diagonals of a parallelogram bisect each other.
4. Find a missing vertex of a parallelogram.

Example: Missing Parallelogram Vertex

A(1, 2), B(5, 3), C(7, 8) and D(x, y) are consecutive vertices of a parallelogram.

The diagonals AC and BD have the same midpoint.

Midpoint of AC:

((1 + 7)/2, (2 + 8)/2) = (4, 5)

Therefore:

(5 + x)/2 = 4

x = 3

(3 + y)/2 = 5

y = 7

D = (3, 7)

Step-by-Step Decision Process

1. If a length is required, use the distance formula.
2. If a point divides a segment in a stated ratio, use the section formula.
3. If the ratio is 1:1, use the midpoint formula.
4. If a point lies on an axis, set the other coordinate to zero.
5. If a point is equidistant from two points, equate their squared distances.
6. If a shape must be identified, calculate only the necessary sides and diagonals.

Common Misconceptions

• Incorrect: Every equidistant point is a midpoint.
  Correct: An equidistant point may lie anywhere on the perpendicular bisector.
• Incorrect: A point on the x-axis has x = 0.
  Correct: A point on the x-axis has y = 0.
• Incorrect: The ratio remains unchanged when endpoints are swapped.
  Correct: Reverse the ratio as well.
• Incorrect: A squared equation can have only one solution.
  Correct: It may produce two valid coordinate values.
• Incorrect: External-division signs are needed.
  Correct: Use the internal formula for the current syllabus.

Coordinate Geometry Solved Examples

Coordinate Geometry solved examples should begin with direct formula questions and progress to unknown-coordinate, axis and geometrical applications.

Level 1: Direct Formula Questions

Example 1: Find the distance between A(1, 2) and B(7, 10).

AB = √[(7 − 1)² + (10 − 2)²]

AB = √(36 + 64)

AB = 10 units

Example 2: Find the midpoint of (−4, 6) and (8, −2).

M = ((−4 + 8)/2, (6 + (−2))/2)

M = (2, 2)

Level 2: Unknown Values and Axis Questions

Example 3: The distance between P(k, 2) and Q(3, 6) is 5 units. Find k.

√[(k − 3)² + (2 − 6)²] = 5

(k − 3)² + 16 = 25

(k − 3)² = 9

k − 3 = ±3

k = 6 or k = 0

Example 4: Find the point dividing A(2, −3) and B(12, 7) in the ratio 3:2.

P = ((3 × 12 + 2 × 2)/5, (3 × 7 + 2 × (−3))/5)

P = (40/5, 15/5)

P = (8, 3)

Level 3: Geometrical Applications

Example 5: Check whether A(0, 0), B(3, 0) and C(0, 4) form a right-angled triangle.

AB² = 9

AC² = 16

BC² = 25

Since AB² + AC² = BC²:

9 + 16 = 25

The triangle is right-angled at A.

Example 6: Check whether A(0, 0), B(2, 3) and C(4, 6) are collinear.

AB = √13

BC = √13

AC = √52 = 2√13

Since AB + BC = AC, the three points are collinear.

Competency-Based Example

A school map represents the library by L(2, 6), the laboratory by S(8, 2) and the assembly ground by G(0, 0).

1. Find the distance between the library and laboratory.
2. Find the midpoint of LS.
3. Decide whether the library or laboratory is closer to the assembly ground.
4. Explain which location would be a fair meeting point for students coming from L and S.

Coordinate Geometry Important Questions with Solutions

Important Coordinate Geometry questions should test direct formulas, reasoning, axis conditions, unknown coordinates and real-life applications.

Formula Warm-Up Questions

1. Find the distance between (2, 3) and (6, 6).
2. Find the distance of (−8, 15) from the origin.
3. Find the midpoint of (−5, 7) and (3, −1).
4. Find the point dividing (1, 4) and (11, 9) in the ratio 2:3.
5. State the coordinates of a point on the y-axis with ordinate −6.

Mathematics Basic Practice

1. Identify the quadrant containing (−4, 7).
2. Find the distance of (5, −2) from the x-axis.
3. Find the midpoint of two locations on a coordinate map.
4. Find the distance between two plotted points.
5. Find a point that divides a route in the ratio 1:2.

Mathematics Standard Practice

1. Find an unknown coordinate from a given distance.
2. Find a point on an axis equidistant from two points.
3. Find the ratio in which a point divides a segment.
4. Verify a triangle or quadrilateral using distances.
5. Find a missing parallelogram vertex using diagonal midpoints.

Sample MCQ

The distance of the point (−7, 4) from the y-axis is:

1. 4 units
2. −7 units
3. 7 units
4. √65 units

Answer: 7 units

The distance from the y-axis is the absolute value of the x-coordinate:

|−7| = 7

Sample Assertion–Reason Question

Assertion: The midpoint of the segment joining (−2, 5) and (6, 1) is (2, 3).

Reason: A midpoint divides a segment internally in the ratio 1:1.

Answer: Both statements are true, and the reason correctly explains the assertion.

Common Mistakes in Coordinate Geometry

Most Coordinate Geometry errors come from incorrect signs, wrong ratio order and using a formula before understanding the question.

• Mistake: Writing a point on the x-axis as (0, y).
  Correction: Write it as (x, 0).
• Mistake: Removing brackets around negative numbers.
  Correction: Write expressions such as (−5 − 3)².
• Mistake: Pairing an x-coordinate with a y-coordinate.
  Correction: Subtract x from x and y from y.
• Mistake: Forgetting the square root.
  Correction: Distance is the positive square root.
• Mistake: Using the midpoint formula for every equidistant point.
  Correction: Check whether the point actually bisects the segment.
• Mistake: Swapping endpoints without reversing the ratio.
  Correction: Reverse both together.
• Mistake: Using external-division signs.
  Correction: Use the internal formula for the current syllabus.
• Mistake: Giving only one square-root value.
  Correction: Check both possible coordinate values.
• Mistake: Omitting units from a distance.
  Correction: Write units unless a specific unit is provided.
• Mistake: Writing coordinates without brackets.
  Correction: Write the final point as (x, y).

How to Present Working in an Exam

1. Given: Write the points and ratio accurately.
2. Formula: State the required formula.
3. Substitution: Use brackets around negative values.
4. Simplification: Show one clear step per line.
5. Conclusion: State the final distance, coordinate or ratio.
6. Check: Confirm the axis, ratio or geometrical condition.

Final-Answer Checklist

• Did I pair x-values with x-values?
• Did I pair y-values with y-values?
• Did I use brackets around negative numbers?
• Did I write the ratio in the correct order?
• Is the internal point between the endpoints?
• Did I include both valid values?
• Did I add units to a distance?
• Did I write coordinates as an ordered pair?

Coordinate Geometry Quick Revision

The fastest revision method is to memorise the four core formulas, practise one question using each formula and review the most common sign and ratio mistakes.

Formula Recap

Distance formula:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

Distance from the origin:

OP = √(x² + y²)

Internal section formula:

P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

Midpoint formula:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)