Coordinate Geometry — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Maths · Ch 7Class 10 · CBSE 2026–27Medium weightage

Coordinate Geometry

What the board actually asks: the Distance Formula (length of a side, type of triangle or quadrilateral, collinearity, equidistant points) and the Section Formula for internal division — including the mid-point and the points of trisection. In the rationalised 2026-27 chapter these two formulas carry essentially every mark; area-of-a-triangle by coordinates has been removed from the syllabus.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

Geometry you can compute.

Coordinate geometry turns points into number-pairs so that lengths and dividing-points become arithmetic. Two tools do all the work: the Distance Formula gives the length between any two points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ , and the Section Formula gives the point that splits the segment $A B$ internally in a chosen ratio $m_{1} : m_{2}$ (with the mid-point as the special $1 : 1$ case). Everything else here — proving a triangle's type, testing whether three points are collinear, finding a missing vertex — is one of these two formulas applied.

Formulae

Distance Formula

Distance formula

P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Tip: The order of subtraction does not matter once squared.

Distance from origin

O P = x^{2} + y^{2}

Tip: Special case with (x₁,y₁) = (0,0).

Section Formula (Internal Division)

Section formula (internal,

m_{1} : m_{2}

)

(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})

Tip: m₁ pairs with the SECOND point's coordinates, m₂ with the first.

Mid-point

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Tip: Section formula with m₁ = m₂.

Collinear Points using the Distance Formula

Three points

A, B, C

are collinear if the sum of two of the distances equals the third, for example

A B + B C = A C

Identifying Shapes using Distances

Triangle: three equal sides is equilateral; two equal sides is isosceles; to test for a right angle, check

a^{2} + b^{2} = c^{2}

numerically on the three distances. Quadrilateral: square = four equal sides and equal diagonals; rhombus = four equal sides but unequal diagonals; rectangle = opposite sides equal and equal diagonals; parallelogram = opposite sides equal (its diagonals share the same mid-point).

Points of Trisection

The two points that divide

A B

into three equal parts divide it in the ratios

1 : 2

and

2 : 1

P (\frac{x _{2} + 2 x _{1}}{3}, \frac{y _{2} + 2 y _{1}}{3})

and

Q (\frac{2 x _{2} + x _{1}}{3}, \frac{2 y _{2} + y _{1}}{3})

Finding the Ratio k : 1

To find the ratio in which a point divides a segment, take the ratio as

k : 1

, put it into the section formula, and solve for

k

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Distance formula

NCERT verbatim — memorise this exact wording· Ch 7,

“So, the distance between the points P(x₁, y₁) and Q(x₂, y₂) is PQ = √((x₂ – x₁)² + (y₂ – y₁)²), which is called the distance formula.”

What it means: The straight-line distance between two points: square the gap in

x

, square the gap in

y

, add them, take the positive square root —

P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

. The order of subtraction doesn't matter because each difference is squared.

Section formula (internal)

NCERT verbatim — memorise this exact wording· Ch 7,

“So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁ : m₂ are ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)). This is known as the section formula.”

What it means: Where a point lands when it splits the segment from

A

B

internally in the ratio

m_{1} : m_{2}

— a weighted average of the endpoints. Watch the cross-pattern: the weight

m_{1}

(the

P A

part) multiplies

x_{2}

, the coordinate of

B

Mid-point formula

NCERT verbatim — memorise this exact wording· Ch 7,

“Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the join of the points A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2).”

What it means: Just average the two

x

's and the two

y

's —

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

. It is the section formula at ratio

1 : 1

Key terms — NCERT verbatim (state word-for-word)

Abscissa (x-coordinate): The distance of a point from the y-axis is called its x-coordinate, or abscissa. (Ch 7, )
Ordinate (y-coordinate): The distance of a point from the x-axis is called its y-coordinate, or ordinate. (Ch 7, )
Distance from the origin: In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by OP = √(x² + y²). (Ch 7, )
Section formula in k : 1 form: If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)). (Ch 7, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Distance between two points

P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

— the length of the segment joining

P (x_{1}, y_{1})

and

Q (x_{2}, y_{2})

. The distance from the origin is the special case

O P = x^{2} + y^{2}

. Read the two differences straight off the coordinates, then square, add, and take the positive root.

FigureThe distance between P(x₁, y₁) and Q(x₂, y₂) is the hypotenuse of a right triangle whose legs are the coordinate differences.

Figure asset pending extraction — the sourced original is Figure.

Horizontal leg = x₂ − x₁, vertical leg = y₂ − y₁; PQ = √(sum of their squares).

2–3 mark “find the distance / length of a side”

The distance formula in disguise — triangles, squares, collinearity

Most questions never say “distance”. To name a triangle, compare its three side-lengths (and check a Pythagoras relation for a right angle); to prove a square, show four equal sides and equal diagonals; to test collinearity, check that the two shorter distances add to the longest (

A B + B C = A C

). All of it is the distance formula, applied a few times.

prove the type of a triangle / quadrilateral, or show three points are collinear (2–3 marks)

Section formula — internal division

The point

P

that splits

A B

internally in the ratio

m_{1} : m_{2}

(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})

. The weights cross over —

m_{1}

multiplies the coordinates of

B

. In the rationalised syllabus only internal division is examined; external division is left for higher classes.

FigureP divides AB internally in the ratio m₁ : m₂ — shown for A(4, −3), B(8, 5), ratio 3 : 1, giving P(7, 3).

Figure asset pending extraction — the sourced original is Figure.

P lies on segment AB with AP : PB = m₁ : m₂; its coordinates are the weighted average of A and B.

find the dividing point, or the ratio — including case-study setups (2–3 marks)

Mid-point & points of trisection

The mid-point is just the average of the coordinates:

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

. The two points of trisection split

A B

into three equal parts — apply the section formula at ratios

1 : 2

and

2 : 1

. A common exam use: a missing vertex of a parallelogram, since its diagonals bisect each other (equal mid-points).

mid-point, points of trisection, or the missing vertex of a parallelogram

Working backwards — find a ratio or a coordinate

When the ratio is unknown, assume it is

k : 1

, apply the section formula, and solve the coordinate you already know. A point on the

y

-axis has

x = 0

; a point on the

x

-axis has

y = 0

— that single fact becomes the equation you solve. Equidistant conditions (

A P = B P

) square to a neat linear relation between

x

and

y

“in what ratio does the y-axis divide…”, find

k

, or an equidistant-point relation

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 7, Example 1,

“Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.”

Apply the distance formula to all three pairs.With

P (3, 2)

Q (- 2, - 3)

R (2, 3)

P Q = 5^{2} + 5^{2} = 50

QR = (- 4)^{2} + (- 6)^{2} = 52

P R = 1^{2} + (- 1)^{2} = 2

.Concept: distance formula

1½ marks

Is it a triangle?The sum of any two of these distances is greater than the third, so

P

Q

R

form a triangle.Concept: triangle inequality

½ mark

Name the type.

P Q^{2} + P R^{2} = 50 + 2 = 52 = Q R^{2}

, so the check

a^{2} + b^{2} = c^{2}

holds numerically and

∠ P = 90°

— it is a right triangle.Concept: right-angle check (numeric)

1 mark

Mark-scheme logic (authored): 3-mark shape: 1½ for the three side-lengths via the distance formula, ½ for the triangle-inequality conclusion, 1 for naming it a right triangle by checking

P Q^{2} + P R^{2} = Q R^{2}

numerically.

Ch 7, Example 3,

“Fig. 7.6 shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer.”

Ch 7, Example 5,

“Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).”

Ch 7, Example 6,

“Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.”

Ch 7, Example 8,

“Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).”

Problem-solving strategies AUTHORED

Need a length or the type of a shape?Distance formula on every relevant pair, then compare. Equal sides, equal diagonals, or a Pythagoras relation names the figure.

Three points — collinear?Compute the three distances; if the two smaller ones sum to the largest, the points are collinear — no area needed.

Split a segment in a ratio?Section formula. Keep the weights straight:

m_{1}

pairs with the FAR point (the cross pattern).

Mid-point or trisection?Mid-point = average the coordinates. Trisection = section formula at

1 : 2

and again at

2 : 1

“In what ratio…” or “find k”?Assume the ratio is

k : 1

, apply the section formula, and use the coordinate you know —

0

on an axis — as the equation to solve.

Common mistakes that cost marks

Where students lose easy marks

Swapping the section-formula weights — writing

m_{1} x_{1}

instead of

m_{1} x_{2}

. The ratio's first part multiplies the SECOND point's coordinate.

Sign slips with negative coordinates —

(x_{2} - x_{1})

with

x_{1}

negative becomes a sum, e.g.

3 - (- 2) = 5

, not

1

Dropping the square root, or taking the negative root — distance is always the positive square root.

Forgetting the axis condition — a point on the

y

-axis has

x = 0

(on the

x

-axis,

y = 0

); that is the equation you solve for the ratio or the point.

Formulae

Distance Formula

Distance formula

P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Tip: The order of subtraction does not matter once squared.

Distance from origin

O P = x^{2} + y^{2}

Tip: Special case with (x₁,y₁) = (0,0).

Section Formula (Internal Division)

Section formula (internal,

m_{1} : m_{2}

)

(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})

Tip: m₁ pairs with the SECOND point's coordinates, m₂ with the first.

Mid-point

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Tip: Section formula with m₁ = m₂.

Collinear Points using the Distance Formula

Three points

A, B, C

are collinear if the sum of two of the distances equals the third, for example

A B + B C = A C

Identifying Shapes using Distances

Triangle: three equal sides is equilateral; two equal sides is isosceles; to test for a right angle, check

a^{2} + b^{2} = c^{2}

Points of Trisection

The two points that divide

A B

into three equal parts divide it in the ratios

1 : 2

and

2 : 1

P (\frac{x _{2} + 2 x _{1}}{3}, \frac{y _{2} + 2 y _{1}}{3})

and

Q (\frac{2 x _{2} + x _{1}}{3}, \frac{2 y _{2} + y _{1}}{3})

Finding the Ratio k : 1

To find the ratio in which a point divides a segment, take the ratio as

k : 1

, put it into the section formula, and solve for

k

Shortcuts & Tricks

Show AB + BC = AC (or any pair summing to the third).

Safe in written answers

The y-axis divides A(x₁,y₁)B(x₂,y₂) in ratio |x₁| : |x₂| (opposite signs); the x-axis in |y₁| : |y₂|.

MCQ / check only: don't write this as working

Other endpoint B = 2M − A when the mid-point M is given.

MCQ / check only: don't write this as working

PA = PB: square to PA² = PB² and avoid roots.

Safe in written answers

Memory Tricks

Antar-antar, square, jod, root.

PQ = √((x₂−x₁)² + (y₂−y₁)²). antar = difference of x and of y; square each; add; root.

Pehla ratio doosre point ke saath.

(m₁x₂ + m₂x₁)/(m₁+m₂). m₁ with x₂ (second point); m₂ with x₁.

Jod ka aadha.

((x₁+x₂)/2, (y₁+y₂)/2). sum of each coordinate, halved.

Mindmap

- - PQ = √((x₂−x₁)² + (y₂−y₁)²)
  - From origin: √(x² + y²)
  - Always the positive root
- - Length of a side
  - Type of triangle / quadrilateral
  - Collinear: AB + BC = AC
  - Equidistant → a relation in x, y
- - P = ((m₁x₂+m₂x₁)/(m₁+m₂), …)
  - Weights cross over
  - k : 1 form to find a ratio
- - Mid-point = average of coordinates
  - Trisection at 1:2 and 2:1
  - Missing vertex of a parallelogram
- - Swapping the section weights
  - Sign slips with negatives
  - Dropping the square root
  - On an axis: x = 0 or y = 0

The Distance Formula — straight from Pythagoras

NCERT builds the distance formula on p.102 by dropping perpendiculars to the x-axis and reading off a right triangle (Fig. 7.5). The two legs come directly from the coordinates, and one application of the Pythagoras theorem gives the length — this is exactly why the formula has that squared-differences shape.

Set up the two points and drop perpendiculars.

P (x_{1}, y_{1}), Q (x_{2}, y_{2})

Draw PR and QS perpendicular to the x-axis; a perpendicular from P meets QS at T (NCERT Fig. 7.5).

Read the horizontal leg from the x-coordinates.

P T = R S = x_{2} - x_{1}

Read the vertical leg from the y-coordinates.

QT = y_{2} - y_{1}

Apply the Pythagoras theorem in right triangle PTQ.

P Q^{2} = P T^{2} + Q T^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Take the positive square root — distance is non-negative.

P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

This is the distance formula.

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

1 markSection A

The distance between two points A and B, on a graph is given as √(10^2 + 7^2). The coordinates of A are (−4, 3). Given that the point B lies in the first quadrant, then all the possible x-coordinates of point B are

(A) multiple of 2
(B) multiple of 3
(C) multiple of 5
(D) multiple of 6

Step-marked solution

[1 mark] AB² = 10² + 7² = 149, and for grid points the only way to write 149 as a sum of two squares is 10² + 7². So for B(x, y), (x + 4)² + (y − 3)² = 149 means the gaps from A(−4, 3) are 10 and 7 in some order: x + 4 = ±10 or ±7. In the first quadrant x > 0, so x = 6 (then y − 3 = 7, y = 10) or x = 3 (then y − 3 = 10, y = 13). Both possible x-coordinates, 3 and 6, are multiples of 3 (3 is not a multiple of 2, 5 or 6). Hence option (b).

4 marksSection E

Raycasting is a technique used in computer games. A snooker game in creation stage is drawn on a coordinate map. It has six pockets P1–P6 and balls including a white ball W(−3, −2) and a green ball G(7, 1). The top rail joining pockets P4 and P6 lies along the line y = 3. (i) How much distance will a ray travel from the green ball to the nearest pocket P4(9, 3)? (ii) A yellow ball is placed at the midpoint of WG. Find its coordinates. (iii) The white ball is hit, rebounds off the rail P4P6 and lands in pocket P2(2, −4). On its way it crosses the x-axis at X(2/7, 0), which divides the path from the striking point on the rail to P2 in the ratio 3 : 4. Find the coordinates of the point on the rail where the ball struck.

Step-marked solution

[1 mark] (i) GP4 = √((9−7)² + (3−1)²) = √(4 + 4) = √8 = 2√2 units.
[1 mark] (ii) Midpoint of W(−3, −2) and G(7, 1) = ((−3+7)/2, (−2+1)/2) = (2, −1/2).
[1 mark] (iii) Let the striking point be S(c, 3). X divides S→P2 in ratio 3 : 4, so x-coordinate: (3·2 + 4·c)/7 = 2/7.
[1 mark] 6 + 4c = 2 ⟹ c = −1 (check y: (3·(−4) + 4·3)/7 = 0 ✓). Striking point = (−1, 3).

3 marksSection C

Three consecutive vertices of a parallelogram ABCD taken in order are A(1, 2), B(4, 3) and C(6, 6). Using the property that the diagonals of a parallelogram bisect each other, find the coordinates of the fourth vertex D.

Step-marked solution

[1 mark] Diagonals of a parallelogram bisect each other ⇒ midpoint of AC = midpoint of BD. Midpoint of AC = ((1 + 6)/2, (2 + 6)/2) = (7/2, 4).
[1 mark] Let D = (x, y). Midpoint of BD = ((4 + x)/2, (3 + y)/2).
[1 mark] (4 + x)/2 = 7/2 ⇒ x = 3; (3 + y)/2 = 4 ⇒ y = 5. Hence D = (3, 5).

Practise Coordinate Geometry competency questions

Read NCERT Chapter 7 on LazyTopper (jemh107)

Official NCERT chapter PDF (jemh107) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 7 (jemh107.pdf, printed pages 99–112). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Distance formula	verbatim	NCERT 2026-27 Ch7
Section formula (internal)	verbatim	NCERT 2026-27 Ch7
Mid-point formula	verbatim	NCERT 2026-27 Ch7 (Special Case)
Abscissa (x-coordinate)	verbatim	NCERT 2026-27 Ch7
Ordinate (y-coordinate)	verbatim	NCERT 2026-27 Ch7
Distance from the origin	verbatim	NCERT 2026-27 Ch7 (Remark 1)
Section formula in k : 1 form	verbatim	NCERT 2026-27 Ch7
Distance formula → type of triangle · NCERT Example 1	example	NCERT 2026-27 Ch7 Example 1 –103
Collinearity via distances · NCERT Example 3	example	NCERT 2026-27 Ch7 Example 3 –104
Equidistant point on the y-axis · NCERT Example 5	example	NCERT 2026-27 Ch7 Example 5 –105
Section formula (internal, 3:1) · NCERT Example 6	example	NCERT 2026-27 Ch7 Example 6
Points of trisection · NCERT Example 8	example	NCERT 2026-27 Ch7 Example 8

Coordinate Geometry · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 7 (jemh107.pdf, printed pages 99–112)

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