Coordinate Geometry – Class 10 Maths Chapter 7 Notes & NCERT PDF

Class 10MathematicsChapter 7NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 7 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 7: Coordinate Geometry

Introduction

In Class IX you learnt to locate points using a pair of coordinate axes: the distance from the y-axis is the x-coordinate (abscissa) and the distance from the x-axis is the y-coordinate (ordinate). Points on the x-axis are (x, 0) and points on the y-axis are (0, y). The chapter begins with a plotting game — joining points like A(4, 8), B(3, 9), … to draw a picture — and reminds you that ax + by + c = 0 is a straight line and y = ax² + bx + c is a parabola. Coordinate geometry is an algebraic tool for studying geometry, used in physics, engineering, navigation, seismology and art.

Key Concepts & Formulas

Distance formula

The distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

It comes from the Pythagoras theorem on the right triangle formed by the horizontal and vertical gaps.

Distance from the origin: OP = √(x² + y²).

Distance is never negative, so take the positive square root. (x₁ − x₂)² = (x₂ − x₁)², so the order does not matter.

Section formula

The point P dividing the join of A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ is

P = ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂))

Mid-point (ratio 1 : 1): ((x₁ + x₂)/2, (y₁ + y₂)/2).

Ratio as k : 1: P = ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)) — useful when the ratio is unknown.

Points of trisection divide a segment in the ratios 1 : 2 and 2 : 1.

Methods Step by Step

Type of triangle: find all three sides; if the sum of the two smaller sides equals the largest, the points are collinear; otherwise check for equal sides or PQ² + PR² = QR² (right angle).

Square / rhombus / rectangle / parallelogram: compare all four sides and both diagonals (square: sides equal and diagonals equal).

Equidistant point: set AP² = BP² and simplify.

Unknown ratio: let the ratio be k : 1, equate one coordinate, solve for k, then check with the other coordinate.

Point on an axis: on the y-axis x = 0; on the x-axis y = 0.

Parallelogram vertex: the diagonals of a parallelogram bisect each other, so the mid-points of AC and BD are the same.

Worked Examples (from the chapter)

Example 1: P(3, 2), Q(−2, −3), R(2, 3): PQ = √50, QR = √52, PR = √2. They form a triangle, and PQ² + PR² = QR², so it is right-angled at P.

Example 2: (1, 7), (4, 2), (−1, −1), (−4, 4): all sides √34 and both diagonals √68 — a square.

Example 3: Desks at A(3, 1), B(6, 4), C(8, 6): AB = 3√2, BC = 2√2, AC = 5√2 and AB + BC = AC, so the students sit in a line.

Example 4: Points equidistant from (7, 1) and (3, 5) satisfy x − y = 2 (the perpendicular bisector).

Example 5: The point on the y-axis equidistant from A(6, 5) and B(−4, 3) is (0, 9).

Example 6: The point dividing (4, −3) and (8, 5) in the ratio 3 : 1 is (7, 3).

Example 7: (−4, 6) divides A(−6, 10) and B(3, −8) in the ratio 2 : 7.

Example 8: Trisection points of A(2, −2) and B(−7, 4) are (−1, 0) and (−4, 2).

Example 9: The y-axis divides (5, −6) and (−1, −4) in the ratio 5 : 1 at (0, −13/3).

Example 10: A(6, 1), B(8, 2), C(9, 4), D(p, 3) is a parallelogram; equating mid-points of AC and BD gives p = 7.

Why the Formulas Work

Deriving the distance formula

From P(x₁, y₁) and Q(x₂, y₂), drop perpendiculars to the x-axis and draw a horizontal line through P.

This forms a right triangle with horizontal side |x₂ − x₁| and vertical side |y₂ − y₁|.

By the Pythagoras theorem, PQ² = (x₂ − x₁)² + (y₂ − y₁)², so PQ = √[(x₂ − x₁)² + (y₂ − y₁)²].

Deriving the section formula

Let P(x, y) divide AB in the ratio m₁ : m₂. Draw lines parallel to the axes through A and P to form two right triangles.

These triangles are similar (AA), so PA/BP = (x − x₁)/(x₂ − x) = m₁/m₂.

Solving, x = (m₁x₂ + m₂x₁)/(m₁ + m₂); in the same way y = (m₁y₂ + m₂y₁)/(m₁ + m₂).

This links Chapter 7 to the similarity ideas of Chapter 6.

Choosing the Right Tool

"How far?" → distance formula.

"Which point divides …?" → section formula.

"In what ratio …?" → take k : 1 and equate one coordinate.

"Centre / mid-point / diagonals bisect" → mid-point formula.

"Lies on an axis" → put the other coordinate equal to 0.

Board Exam Focus

Direct use of the distance and section formulas, collinearity tests, and "find the ratio" problems are asked every year.

Show every substitution clearly; state the formula first.

Formula Sheet

Distance: √[(x₂ − x₁)² + (y₂ − y₁)²]

Distance from origin: √(x² + y²)

Section (m₁ : m₂): ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂))

Mid-point: ((x₁ + x₂)/2, (y₁ + y₂)/2)

Point on x-axis: (x, 0); on y-axis: (0, y)

Common Mistakes

Swapping m₁ and m₂ in the section formula (m₁ goes with the second point B).

Forgetting to square negative differences correctly: (−2 − 3)² = 25, not −25.

Leaving answers as √50 when a simplified form 5√2 is needed for comparison.

Using AB + BC = AC with rounded decimals instead of exact surds.

For a parallelogram, assuming sides are equal instead of using bisecting diagonals.

💡 Key Learning Points

✓The distance between two points is √[(x₂ − x₁)² + (y₂ − y₁)²], derived from the Pythagoras theorem.
✓The section formula gives the point dividing a segment internally in the ratio m₁ : m₂; the mid-point is the special case 1 : 1.
✓Distances help decide whether points are collinear or form a right, isosceles or equilateral triangle, a square or a parallelogram.
✓An unknown ratio is best found by taking it as k : 1 and equating one coordinate.
✓Coordinate geometry links algebra and geometry: equidistant points give linear equations such as x − y = 2.

👨‍🏫 Teaching Tips

→Begin with the book's plotting game (A(4, 8) to L(5, 8)) so students enjoy locating points before new formulas.
→Derive the distance formula on graph paper by drawing the right triangle between two points.
→Use the classroom-desk example (Ashima, Bharti, Camella) with actual desks and a coordinate grid on the floor.
→Show both ways of finding a ratio (m₁ : m₂ and k : 1) and let students choose the faster one.
→Have students verify the square in Example 2 in two ways: equal sides + equal diagonals, and the Pythagoras check at one vertex.
→Give a 'find the missing vertex' puzzle using mid-points of diagonals of a parallelogram.
→Encourage exact surd answers and a final check by substituting back.

📋 Assessment Questions

Find the distance between (2, 3) and (5, 7). (Answer: 5)
Show that (3, 1), (6, 4) and (8, 6) are collinear.
Find the point dividing the join of (4, −3) and (8, 5) in the ratio 3 : 1.
In what ratio does the y-axis divide the join of (5, −6) and (−1, −4)?
Find p if (6, 1), (8, 2), (9, 4) and (p, 3) are vertices of a parallelogram in order.

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