CBSE Class 10 Mathematics Coordinate Geometry Notes Set 11

Cartesian Coordinates

The position of a point in a plane is fixed by its distances from two perpendicular lines, the x-axis and the y-axis, meeting at the origin O. A point P(x, y) has abscissa x (distance from the y-axis) and ordinate y (distance from the x-axis). The axes divide the plane into four quadrants: I (+, +), II (−, +), III (−, −) and IV (+, −). Points on the x-axis have y = 0 and points on the y-axis have x = 0.

Polar Coordinates

A point can also be given as (r, θ), where r = OP is the distance from the origin (pole) and θ is the angle OP makes with the positive x-axis, measured anticlockwise.
1. From polar to Cartesian: x = r cos θ and y = r sin θ.
2. From Cartesian to polar: r = √(x² + y²) and tan θ = y/x.
Example: (4, 4) has r = 4√2 and θ = 45°, so its polar form is (4√2, 45°). The point (6, 60°) is (6 cos 60°, 6 sin 60°) = (3, 3√3).

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 from the origin is \( \sqrt{x^2 + y^2} \). It follows from the Pythagoras theorem applied to the right triangle with horizontal side |x₂ − x₁| and vertical side |y₂ − y₁|.
Uses: three points are collinear if the largest of the three distances equals the sum of the other two; a triangle is isosceles, equilateral or right-angled by comparing the squares of its sides; a point is on a circle with a given centre if its distance from the centre equals the radius.
Example: the distance between (3, 4) and (−5, −2) is √(64 + 36) = 10.

Section Formula

The point P dividing the join of A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n has coordinates \( \left(\frac{mx_2 + nx_1}{m + n},\ \frac{my_2 + ny_1}{m + n}\right) \). For external division replace n by −n: \( \left(\frac{mx_2 - nx_1}{m - n},\ \frac{my_2 - ny_1}{m - n}\right) \).
Mid-point (m = n): \( \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right) \).
Example: the points of trisection of (2, −1) and (5, 2) are (3, 0) and (4, 1).
To find an end point when the other end and the dividing point are known, put the unknown as (h, k) in the formula and solve.

Area of a Triangle

The area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is \( \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right| \). The three points are collinear if this area is 0. For a polygon list the vertices in order and use the shoelace sum divided by 2. The triangle formed by the mid-points of the sides of a triangle has one quarter of its area.

Straight Lines and Circles

1. Slope of the line through two points: m = (y₂ − y₁)/(x₂ − x₁). Parallel lines have equal slopes; perpendicular lines have slopes whose product is −1.
2. Slope-point form: y − y₁ = m(x − x₁). Intercept form: x/a + y/b = 1.
3. Locus: the path of a point that obeys a stated condition. The points equidistant from two fixed points lie on the perpendicular bisector of their join.
4. Circle with centre (h, k) and radius r: (x − h)² + (y − k)² = r².