Coordinate Geometry — Class 10 Mathematics

What CBSE examines here (2026-27). Two things: the distance formula and the section formula for internal division (with the midpoint as its special case). The chapter runs to two exercises only, 7.1 and 7.2. The area of a triangle from coordinates was removed, and with it the old "area = 0" test for collinearity — use distances instead. The area formula and the centroid are in the appendix, marked as background.

"Geometry and algebra unite — a point becomes a pair of numbers, a line becomes an equation."

1. About the Chapter

Coordinate Geometry (or Cartesian Geometry) describes geometric shapes using algebra. Founded by René Descartes (1637) and earlier influenced by Indian mathematicians.

Key Idea

Every point in the plane is identified by (x, y) — ordered pair of coordinates. Once a shape's vertices are numbers, you can prove things about it by arithmetic: that a quadrilateral is a square, that three friends are sitting in a straight line, that a point lies on the y-axis.

What This Chapter Covers

• Recap of the Cartesian plane
• Distance formula — and the classification problems it makes possible
• Section formula (internal division), including the midpoint
• Applications

2. The Cartesian Plane (Quick Review)

Axes

• x-axis: horizontal
• y-axis: vertical
• They meet at ORIGIN (0, 0)

Quadrants

• Q1: top-right, (+x, +y)
• Q2: top-left, (−x, +y)
• Q3: bottom-left, (−x, −y)
• Q4: bottom-right, (+x, −y)

Plot a Point

P(3, 5): go 3 units right, 5 units up. Q(−2, −4): go 2 units left, 4 units down.

3. Distance Formula

Statement

Distance between P(x₁, y₁) and Q(x₂, y₂):

PQ = √((x₂−x₁)² + (y₂−y₁)²)

Derivation

Form right triangle: horizontal leg = (x₂−x₁), vertical leg = (y₂−y₁). Apply Pythagoras.

Examples

Example 1: Find distance between A(2, 3) and B(5, 7).

• PQ = √((5−2)² + (7−3)²) = √(9 + 16) = √25 = 5 units

Example 2: Find distance from origin (0,0) to P(−3, 4).

• = √(9 + 16) = √25 = 5 units

Example 3: Verify (0,0), (3,4), (−3,4) form an isosceles triangle.

• d₁ from (0,0) to (3,4): 5
• d₂ from (0,0) to (−3,4): 5
• d₃ from (3,4) to (−3,4): 6
• Two sides equal (5 = 5), so isosceles ✓

4. Section Formula (Internal Division)

Statement

If P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in ratio m:n (internally):

x = (mx₂ + nx₁) / (m + n) y = (my₂ + ny₁) / (m + n)

Midpoint (Special Case)

When P is the midpoint, m = n = 1: Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)

Examples

Example 1: Find point dividing line joining A(2, 3) and B(8, 9) in ratio 2:1.

• x = (2·8 + 1·2)/3 = 18/3 = 6
• y = (2·9 + 1·3)/3 = 21/3 = 7
• Point: (6, 7)

Example 2: Find midpoint of A(4, −3) and B(8, 5).

• ((4+8)/2, (−3+5)/2) = (6, 1)

5. Collinearity and Classifying Shapes

With only the distance formula you can settle a surprising number of questions.

Are three points collinear?

Compute all three pairwise distances. If the two smaller ones add up to the largest, the points lie on one straight line; if their sum is larger, they form a triangle.

Example: A(3, 1), B(6, 4), C(8, 6).

• AB = √(9 + 9) = 3√2 ≈ 4.24
• BC = √(4 + 4) = 2√2 ≈ 2.83
• AC = √(25 + 25) = 5√2 ≈ 7.07

AB + BC = 3√2 + 2√2 = 5√2 = AC exactly, so A, B and C are collinear. This is the textbook's own classroom-desks example.

Why this works. Going from A to C via B can never be shorter than going straight. Equality happens only when the detour is not a detour at all — that is, when B is on the segment AC.

What kind of quadrilateral is it?

Find all four sides and both diagonals, taking the vertices in the order given. Then:

The diagonals are what people forget — four equal sides alone only gets you as far as a rhombus.

Check the points really form a quadrilateral first. In Exercise 7.1 Q6(ii), three of the four given points turn out to be collinear, so there is no quadrilateral at all. The question says "if any" for exactly that reason.

6. Applications

Coordinate Geometry in Various Problems

Type 1: Show a point lies on an axis — a point on the x-axis is (x, 0), on the y-axis it is (0, y). Substitute the known zero and you have one unknown, not two.

Type 2: Find a point equidistant from two others — set the two squared distances equal. The squares cancel and what remains is linear.

Type 3: Find the centre of a circle through given points — the centre is equidistant from all of them, so apply Type 2 twice.

Type 4: Prove a quadrilateral is a parallelogram — show its diagonals bisect each other, i.e. the two midpoints coincide.

Type 5: Find where a segment meets an axis — use the k : 1 form of the section formula and set the relevant coordinate to zero.

7. Worked Examples

Example 1: Equilateral Triangle

Show that points (0, 0), (3, √3), and (3, −√3) form an equilateral triangle.

• Side 1: from (0,0) to (3, √3) = √(9 + 3) = √12 = 2√3
• Side 2: from (0,0) to (3, −√3) = √(9 + 3) = 2√3
• Side 3: from (3, √3) to (3, −√3) = √(0 + 12) = 2√3
• All sides equal = 2√3 → EQUILATERAL ✓

Example 2: Find Coordinates

Find coordinates of points trisecting line joining A(2, 1) and B(5, 8).

• One trisecting point divides 1:2:
  • x = (1·5 + 2·2)/3 = 9/3 = 3
  • y = (1·8 + 2·1)/3 = 10/3
  • Point: (3, 10/3)
• Other trisecting point divides 2:1:
  • x = (2·5 + 1·2)/3 = 12/3 = 4
  • y = (2·8 + 1·1)/3 = 17/3
  • Point: (4, 17/3)

Example 3: Ratio in which an axis cuts a segment

Find the ratio in which the x-axis divides the segment joining A(1, −5) and B(−4, 5).

• Let the ratio be k : 1. Then the y-coordinate of the dividing point is (5k − 5)/(k + 1).
• On the x-axis the y-coordinate is 0, so 5k − 5 = 0, giving k = 1.
• The ratio is 1 : 1 — the x-axis cuts AB at its midpoint, which is (−3/2, 0).

Using the k : 1 form rather than m₁ : m₂ is what keeps this to one unknown.

Example 4: Circle Centre

A circle passes through (0, 0), (2, 0), and (0, 2). Find centre.

• Centre is equidistant from all three points.
• Let centre = (h, k). Distance to each = r.
• h² + k² = (h−2)² + k² → 4h = 4 → h = 1
• h² + k² = h² + (k−2)² → 4k = 4 → k = 1
• Centre: (1, 1)

8. Common Mistakes

1. Wrong coordinate order

   • Always (x, y) — first horizontal, then vertical.

2. Swapping m₁ and m₂ in the section formula

   • m₁ : m₂ means PA : PB, so m₁ multiplies x₂ — the coordinate of the far point. Reversing them gives the point that divides in the opposite ratio, which is a different point entirely.

3. Calling a shape a square on four equal sides alone

   • Four equal sides gives a rhombus. You need the diagonals equal as well before you may say square.

4. Testing collinearity with the area formula

   • That formula is no longer part of this chapter. Compare the three distances: collinear exactly when the two smaller add to the largest.

5. Losing a root when a distance condition gives a quadratic

   • (y + 3)² = 36 has two answers, y = 3 and y = −9. Both are usually valid; say so.

6. Wrong midpoint formula

   • Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2). Sum first, then divide.

9. Indian Heritage

While Cartesian coordinates were formalised by Descartes (1637), Indian mathematicians used:

• Bhaskara II (12th century): worked with curves and surfaces
• Vedic geometry: precise altar geometry
• Aryabhata (5th century): used coordinate-like thinking

Modern coordinate geometry combines Indian algebraic ideas with European geometric notation.

10. Conclusion

Coordinate Geometry bridges algebra and geometry:

• Points become numerical pairs
• Lines become equations
• Geometric problems solved algebraically

Master two formulas and the problems they unlock:

• the distance formula, and using it to classify triangles and quadrilaterals and to test collinearity
• the section formula, in both the m₁ : m₂ and the k : 1 form

Foundation for:

• Class 11 Conic Sections
• Class 12 3D Geometry, Calculus
• Engineering, Physics
• Computer Graphics

The Cartesian Plane is the canvas on which all of mathematics is drawn.

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. The area of a triangle from coordinates was removed from this chapter when the syllabus was rationalised, taking old Exercises 7.3 and 7.4 with it. The rationalised chapter runs 7.1 Introduction → 7.2 Distance Formula → 7.3 Section Formula → 7.4 Summary, and the summary lists four results, none about area. It is kept here because guidebooks still lead with it and because it is genuinely useful later.

Area of a triangle from its vertices

For a triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):

$\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

The absolute value is needed because the expression inside comes out negative when the vertices are taken clockwise; area is not.

Example. A(1, 2), B(4, 5), C(6, 1).

Area = ½ |1(5 − 1) + 4(1 − 2) + 6(2 − 5)| = ½ |4 − 4 − 18| = ½ × 18 = 9 square units

The old collinearity test. If the three points lie on one line the "triangle" is degenerate and the formula returns zero. For A(2, 4), B(4, 6), C(6, 8):

Area = ½ |2(6 − 8) + 4(8 − 4) + 6(4 − 6)| = ½ |−4 + 16 − 12| = 0, so the points are collinear.

Do not use this in the exam. Section 5 of this chapter gives the method that is still in the syllabus — compare the three distances. It is the one the textbook itself uses.

Centroid of a triangle

The centroid, where the three medians meet, is the plain average of the vertices:

$G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

For (0, 0), (6, 0) and (3, 9) this gives G = (3, 3). The centroid is not part of the rationalised Class 10 chapter at all.

External division

The chapter's closing Note to the Reader points out that when P lies on the line AB but outside the segment, with PA : PB = m₁ : m₂, we say P divides AB externally — and states plainly that the formula for that case comes in higher classes. So you are told it exists and told not to worry about it yet.