Lesson 1 of 10 · 7 min

Coordinate geometry recap

NCERT §9.1

Riya is drawing a plan of an imaginary town on graph paper for a project. The town hall sits at the origin, a well at W(4, −1), and a straight canal runs across the sheet. Before she can say anything about the canal, she needs the tools that measure points. Which ones?

The story this chapter follows: Riya's town on graph paper

Imagine Riya, who is drawing the plan of a made-up town on graph paper. The town hall sits at the origin, a well at W(4, −1), a straight canal runs along 3x − 4y − 26 = 0, and later a road runs beside it. Each square is one unit. Every question about the town, from the steepness of the canal to the length of a pipe from the well, turns into a question about straight lines.

The lesson in notes

In short

Coordinate geometry studies figures through algebra: each point of the plane is a pair (x, y), and a curve becomes an equation. René Descartes set out this method in his book La Géométrie of 1637.

The point (6, −4) lies 6 units from the y-axis on the positive x side and 4 units below the x-axis. The point (3, 0) sits on the x-axis itself.

Distance formula: PQ = √((x₂ − x₁)² + (y₂ − y₁)²). From (6, −4) to (3, 0) it is √(9 + 16) = 5.

Section formula: the point dividing the join of (x₁, y₁) and (x₂, y₂) internally in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). For A(1, −3) and B(−3, 9) in the ratio 1 : 3 it is the origin (0, 0). With m = n it gives the mid-point, here (−1, 3).

Area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. For (4, 4), (3, −2) and (−3, 16) it is ½|−72 + 36 − 18| = 27.

If that area comes out zero, the three points lie on one line; they are collinear. The points (1, 1), (2, 3) and (4, 7) give area 0.