CBSE Class 10 Mathematics Coordinate Geometry Notes Set 15

Cartesian Plane Basics

The position of a point is given by an ordered pair (x, y): x is the abscissa (distance from the y-axis) and y is the ordinate (distance from the x-axis). The axes divide the plane into four quadrants: I (+, +), II (−, +), III (−, −), IV (+, −).
A point on the x-axis has y = 0, i.e. (x, 0), and a point on the y-axis has x = 0, i.e. (0, y). The origin is (0, 0).

Distance Formula

The distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = \( \sqrt{(x_2 − x_1)^2 + (y_2 − y_1)^2} \).
The distance of P(x, y) from the origin is \( \sqrt{x^2 + y^2} \).
Uses: to show that three points are collinear (the sum of two distances equals the third), to identify types of triangles and quadrilaterals (equal sides, Pythagoras for right angles, equal diagonals for a rectangle), and to find a point equidistant from two given points.

Section Formula

The point P dividing the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n has coordinates
P = \( \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \).
Midpoint (m = n = 1): ((x₁ + x₂)/2, (y₁ + y₂)/2).
Centroid of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃): ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).
A line cuts the y-axis (x = 0) in the ratio x₁ : x₂ in magnitude (so that x₁ and x₂ have opposite signs), and the x-axis in the ratio of the absolute y-coordinates.

Area of a Triangle

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
Three points are collinear exactly when this area is zero.
The area of a quadrilateral can be found by splitting it along a diagonal into two triangles, or by the shoelace method.

Useful Properties

1. The diagonals of a parallelogram bisect each other, so the fourth vertex D of parallelogram ABCD is A + C − B.
2. The diagonals of a rectangle are equal; those of a rhombus are perpendicular and bisect each other; those of a square are equal and perpendicular.
3. The centre of a circle passing through three points is found by equating the squared distances from the centre to those points.
4. The circumcentre of a right triangle is the midpoint of its hypotenuse.

Worked Examples

Example 1. Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
Solution: Let the point be (x, 0). (x − 2)² + 25 = (x + 2)² + 81, so −8x = 56 and x = −7. The point is (−7, 0).

Example 2. Find the ratio in which the point (−3, k) divides the segment joining (−5, −4) and (−2, 3). Find k.
Solution: Let the ratio be m : n. x: (−2m − 5n)/(m + n) = −3 gives m = 2n, so the ratio is 2 : 1. Then k = (2 × 3 + 1 × (−4))/3 = 2/3.

Example 3. Find the area of the triangle with vertices (1, −1), (−4, 6) and (−3, −5).
Solution: Area = ½ |1(6 + 5) + (−4)(−5 + 1) + (−3)(−1 − 6)| = ½ |11 + 16 + 21| = 24 square units.

Example 4. Find k if the points (k, 2 − 2k), (−k + 1, 2k) and (−4 − k, 6 − 2k) are collinear.
Solution: Setting the area to zero gives k = −1 or k = 1/2.

Example 5. The vertices of a triangle are (1, 2), (4, 6) and (7, 2). Find the centroid.
Solution: ((1 + 4 + 7)/3, (2 + 6 + 2)/3) = (4, 10/3).