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Chapter 7: Coordinate Geometry

Descartes' great idea was to label every point in the plane by a pair of numbers (x,y)(x, y). Geometry then becomes algebra: distances, midpoints, intersections , all computable from coordinates. This chapter delivers the two formulas you'll use forever after: the distance formula and the section formula, with the midpoint formula as a corollary.

For board exams: 11-mark MCQs on distance/midpoint, 22- and 33-mark questions on section formula, collinearity tests, and a 44- to 55-mark problem about identifying a shape (parallelogram, square, rhombus) from given vertices, or finding a point that divides a line segment in a given ratio.

Coordinate geometry is the gateway to higher mathematics. The same ideas , distance, midpoint, ratio , extend to three dimensions, to vectors, to complex numbers, to fields most students meet in calculus. Master them now.

What's inside

  • Coordinate plane and axes , quick refresher.
  • Distance formula , derivation via Pythagoras.
  • Section formula , internal division in a given ratio, with midpoint as a special case.
  • Collinearity and area , testing whether three points are collinear.
  • Applications and shape recognition , quadrilaterals from vertices.

Key results / Formula card

ResultFormula
DistancePQ=(x2−x1)2+(y2−y1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Section formula (internal, ratio m:nm : n)(mx2+nx1m+n, my2+ny1m+n)\left(\dfrac{m x_2 + n x_1}{m + n},\ \dfrac{m y_2 + n y_1}{m + n}\right)
Midpoint(x1+x22, y1+y22)\left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)
Collinearity (area = 0)x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0
Distance from originx2+y2\sqrt{x^2 + y^2}

Sub-topics

5 pages
01
The Cartesian plane , a refresher
Axes, quadrants, coordinates, and a quick recap before deriving the distance and section formulas.
02
The distance formula
Derivation of the distance between two points in the plane, and key applications.
03
The section formula and midpoint
Find the coordinates of a point that divides a line segment in a given ratio; the midpoint is the special case 1:1.
04
Collinearity and area of a triangle
Compute the area of a triangle from its vertices and use the area-zero test for collinear points.
05
Applications and shape recognition
Classify quadrilaterals (parallelogram, rhombus, rectangle, square) and recognise triangles (isoceles, equilateral, right) from coordinates.

Practice quiz

Answer the questions; explanations appear after each.

Quiz
Chapter 7 : Mixed practice
10 questions · pick the best answer
Q1

Distance between (3,−2)(3, -2) and (−1,4)(-1, 4):

Q2

Midpoint of (2,7)(2, 7) and (−4,5)(-4, 5):

Q3

Point dividing (1,2)(1, 2) to (6,7)(6, 7) in 2:32:3:

Q4

Area of triangle (1,1),(4,3),(2,5)(1,1), (4,3), (2,5):

Q5

Centroid of (2,3),(−1,4),(5,−1)(2,3), (-1,4), (5,-1):

Q6

Points (1,4),(3,1),(−1,7)(1, 4), (3, 1), (-1, 7):

Q7

Distance from origin to (7,−24)(7, -24):

Q8

yy-axis divides segment from (−3,5)(-3, 5) to (4,−2)(4, -2) in ratio:

Q9

Mid-point of ABAB is (2,−3)(2, -3) and A=(5,1)A = (5, 1). Then BB is:

Q10

Square's diagonal length for side aa: