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CBSE Class 10 · Maths · board exam 2027

Chapter 7: Coordinate Geometry — notes

The chapter’s 5 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.

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  1. 1. Coordinates of a point

    Look at the figure.

    • • x-axis (horizontal), y-axis (vertical)
    • • Origin O = (0, 0)
    • • P(7, 3)
    • • x-coordinate (abscissa) = 7 = distance from the y-axis
    • • y-coordinate (ordinate) = 3 = distance from the x-axis
    • • Q(−3, 4): 3 units left of the y-axis, 4 units above the x-axis
    • • Distance of Q from the y-axis = 3 (never negative)
    • • On the x-axis: (x, 0)
    • • On the y-axis: (0, y)
    • • (2, −6): abscissa = 2, ordinate = −6
    • • 2 − (−6) = 2 + 6 = 8

  2. 2. The distance formula

    Look at the figure.

    • • B: 36 km east and 15 km north of A
    • • AB² = 36² + 15²
    • • AB² = 1296 + 225 = 1521
    • • AB = √1521 = 39 km
    • • P(x₁, y₁), Q(x₂, y₂)
    • • PT ∥ x-axis, QT ∥ y-axis, ∠T = 90°
    • • PT = x₂ − x₁, QT = y₂ − y₁
    • • PQ² = PT² + QT²
    • • PQ² = (x₂ − x₁)² + (y₂ − y₁)²
    • • PQ = √((x₂ − x₁)² + (y₂ − y₁)²)
    • • OP = √(x² + y²)
    • • Example: P(3, 2), Q(−2, −3)
    • • PQ = √((x₂ − x₁)² + (y₂ − y₁)²)
    • • x₁ = 3, y₁ = 2, x₂ = −2, y₂ = −3
    • • x₂ − x₁ = −2 − 3 = −5
    • • y₂ − y₁ = −3 − 2 = −5
    • • (−5)² = 25, (−5)² = 25
    • • 25 + 25 = 50
    • • PQ = √50 = √(25 × 2) = 5√2 units
    • • (x₁ − x₂)² = (x₂ − x₁)² ⇒ same distance either way
    • • R(2, 3): QR² = (2 + 2)² + (3 + 3)² = 16 + 36 = 52
    • • PR² = (2 − 3)² + (3 − 2)² = 1 + 1 = 2
    • • PQ² + PR² = 50 + 2 = 52 = QR² ⇒ ∠P = 90°

  3. 3. Using the distance formula

    Look at the figure.

    • • A(1, 7), B(4, 2), C(−1, −1), D(−4, 4): a square?
    • • Square: 4 sides equal and 2 diagonals equal
    • • AB² = (4 − 1)² + (2 − 7)² = 9 + 25 = 34
    • • BC² = (−1 − 4)² + (−1 − 2)² = 25 + 9 = 34
    • • CD² = (−4 + 1)² + (4 + 1)² = 9 + 25 = 34
    • • DA² = (1 + 4)² + (7 − 4)² = 25 + 9 = 34
    • • AC² = (−1 − 1)² + (−1 − 7)² = 4 + 64 = 68
    • • BD² = (−4 − 4)² + (4 − 2)² = 64 + 4 = 68
    • • ⇒ ABCD is a square
    • • A(3, 1), B(6, 4), C(8, 6): collinear?
    • • AB = √(9 + 9) = √18 = 3√2
    • • BC = √(4 + 4) = √8 = 2√2
    • • AC = √(25 + 25) = √50 = 5√2
    • • AB + BC = 3√2 + 2√2 = 5√2 = AC
    • • AB + BC = AC ⇒ A, B, C are collinear
    • • P(x, y) equidistant from A(7, 1) and B(3, 5)
    • • PA = PB ⇒ PA² = PB²
    • • (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)²
    • • x² − 14x + 49 + y² − 2y + 1 = x² − 6x + 9 + y² − 10y + 25
    • • −14x − 2y + 50 = −6x − 10y + 34
    • • −8x + 8y = −16
    • • x − y = 2
    • • P(0, y) on the y-axis; A(6, 5), B(−4, 3)
    • • (0 − 6)² + (y − 5)² = (0 + 4)² + (y − 3)²
    • • 36 + y² − 10y + 25 = 16 + y² − 6y + 9
    • • 61 − 10y = 25 − 6y
    • • 36 = 4y ⇒ y = 9
    • • P = (0, 9)

  4. 4. The section formula

    Look at the figure.

    • • AP : PB = 1 : 2
    • • A(0, 0), B(36, 15) ⇒ P(12, 5)
    • • A(x₁, y₁), B(x₂, y₂), P(x, y), AP : PB = m₁ : m₂
    • • Two right triangles similar (AA)
    • • m₁/m₂ = (x − x₁)/(x₂ − x)
    • • m₁(x₂ − x) = m₂(x − x₁)
    • • m₁x₂ − m₁x = m₂x − m₂x₁
    • • m₁x₂ + m₂x₁ = (m₁ + m₂)x
    • • x = (m₁x₂ + m₂x₁)/(m₁ + m₂), y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
    • • Example: A(4, −3), B(8, 5), ratio 3 : 1
    • • x₁ = 4, y₁ = −3, x₂ = 8, y₂ = 5, m₁ = 3, m₂ = 1
    • • x = (3 × 8 + 1 × 4)/(3 + 1) = (24 + 4)/4 = 28/4 = 7
    • • y = (3 × 5 + 1 × (−3))/(3 + 1) = (15 − 3)/4 = 12/4 = 3
    • • Point = (7, 3)
    • • Ratio k : 1: x = (kx₂ + x₁)/(k + 1), y = (ky₂ + y₁)/(k + 1)
    • • Mid-point = ((x₁ + x₂)/2, (y₁ + y₂)/2)
    • • Trisect A(2, −2), B(−7, 4): AP = PQ = QB
    • • P: AP : PB = 1 : 2
    • • x = (1 × (−7) + 2 × 2)/(1 + 2) = (−7 + 4)/3 = (−3)/3 = −1
    • • y = (1 × 4 + 2 × (−2))/(1 + 2) = (4 − 4)/3 = 0
    • • P = (−1, 0)
    • • Q: AQ : QB = 2 : 1
    • • x = (2 × (−7) + 1 × 2)/(2 + 1) = (−14 + 2)/3 = (−12)/3 = −4
    • • y = (2 × 4 + 1 × (−2))/(2 + 1) = (8 − 2)/3 = 6/3 = 2
    • • Q = (−4, 2)

  5. 5. Finding the ratio, the axes and the parallelogram

    Look at the figure.

    • • (−4, 6) divides A(−6, 10), B(3, −8) in k : 1
    • • x = (k × 3 + (−6))/(k + 1) = (3k − 6)/(k + 1)
    • • (3k − 6)/(k + 1) = −4
    • • 3k − 6 = −4k − 4
    • • 3k + 4k = −4 + 6
    • • 7k = 2 ⇒ k = 2/7
    • • Ratio = 2 : 7
    • • y = (−8k + 10)/(k + 1): −16/7 + 10 = 54/7, 2/7 + 1 = 9/7
    • • y = 54/7 ÷ 9/7 = 6 ✓
    • • y-axis divides A(5, −6), B(−1, −4) in k : 1
    • • On the y-axis: x = 0
    • • x = (k × (−1) + 5)/(k + 1) = (−k + 5)/(k + 1) = 0
    • • −k + 5 = 0 ⇒ k = 5
    • • Ratio = 5 : 1
    • • y = (5 × (−4) + 1 × (−6))/(5 + 1) = (−26)/6 = −13/3
    • • Point = (0, −13/3)
    • • x-axis: put y = 0; y-axis: put x = 0
    • • Parallelogram ABCD: A(6, 1), B(8, 2), C(9, 4), D(p, 3)
    • • Mid-point of AC = Mid-point of BD
    • • Mid-point of AC = ((6 + 9)/2, (1 + 4)/2) = (15/2, 5/2)
    • • Mid-point of BD = ((8 + p)/2, (2 + 3)/2) = ((8 + p)/2, 5/2)
    • • 8 + p = 15 ⇒ p = 7

Summary

Very good! We started with coordinates: the abscissa and the ordinate, and the forms x comma zero on the x-axis and zero comma y on the y-axis. Then the Pythagoras theorem gave us the distance formula. We used it to check squares, collinear points and right triangles, and to find a point equidistant from two points. Then the section formula, with m one going with x two, the mid-point formula as its 1 is to 1 case, the points of trisection, and the ratio in which an axis divides a segment. Write the two formulas in your copy once without looking. Your homework is on your screen. If anything was not clear, press the Ask button and ask me. Done today!

AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 7. Written from the textbook, checked twice.

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