Coordinate Geometry

1. Chapter at a glance

• To locate a point on a plane, a pair of perpendicular coordinate axes is used; x-coordinate (abscissa) is distance from y-axis and y-coordinate (ordinate) is distance from x-axis.
• Distance between any two points P(x₁, y₁) and Q(x₂, y₂) is found by applying Pythagoras theorem after drawing perpendiculars to the axes.
• A point dividing the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ has coordinates obtained via similarity of triangles (AA criterion).
• Special case of section formula gives the mid-point when the ratio is 1 : 1.
• Distance formula and section formula together allow verification of collinearity, type of triangle/quadrilateral, and location of points equidistant from given points.
• Points on x-axis are of form (x, 0) and on y-axis of form (0, y); distance from origin is √(x² + y²).
• Properties such as diagonals of a parallelogram bisecting each other or equal sides + equal diagonals confirming a square follow directly from these formulas.

2. Definitions, theorems and results

• Distance formula: The distance between points P(x₁, y₁) and Q(x₂, y₂) is √[(x₂ – x₁)² + (y₂ – y₁)²]. (Derived via Pythagoras theorem on right triangle formed by perpendiculars to axes; distance is non-negative, hence only positive square root is taken.)
• Section formula (internal division): The coordinates of the point P(x, y) which divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)). (Derived via AA similarity of triangles formed by perpendiculars to axes.)
• Mid-point formula: The mid-point of the line segment joining A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). (Special case of section formula when m₁ : m₂ = 1 : 1.)
• Distance of point P(x, y) from origin O(0, 0) is √(x² + y²).
• No separate theorems/axioms/ lemmas are stated beyond the derivations above; Pythagoras theorem and AA similarity criterion (from Class IX/VI) are used as tools.

(Note: NCERT presents derivations of both formulas; students are expected to reproduce the distance formula derivation using Pythagoras in exams.)

3. Formula sheet

4. Solved-example patterns

• Collinearity check: Compute distances AB, BC, AC using distance formula; verify if AB + BC = AC (or any permutation).
• Type of triangle/quadrilateral: Compute all side lengths and diagonals; apply converse of Pythagoras for right angle, check equal sides + equal diagonals for square, or equal sides for rhombus/isosceles.
• Equidistant point: Set distances equal (AP = BP), simplify the resulting equation in x and y; for point on axis, substitute x = 0 or y = 0 directly.
• Section formula application: Substitute given points and ratio directly into section formula to find coordinates; or solve for unknown ratio k by equating one coordinate and verifying with the other.
• Trisection / multiple division points: Apply section formula successively with ratios 1:2 and 2:1 (or equivalent).
• Parallelogram vertices: Use that diagonals bisect each other → equate mid-points of both diagonals and solve for unknown coordinate.
• Point dividing in given ratio on axis: Set abscissa or ordinate to zero in section formula and solve for ratio k.

5. Common mistakes and exam pitfalls

• Forgetting that distance is always the positive square root (never write negative).
• Sign errors when subtracting coordinates (x₂ – x₁ vs x₁ – x₂) — squares make it same, but students often mishandle in section formula.
• Writing section formula with m₁ and m₂ swapped (remember PA : PB = m₁ : m₂).
• Assuming external division when question says “internally”; text covers only internal case.
• Missing the verification step after finding ratio k (must check both coordinates).
• Forgetting to confirm collinearity or Pythagoras converse after distance calculations.
• Units omitted or wrong assumption that points lie on axes when they do not.