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IB DP Maths AI HL Study Notes

3.1.1 Coordinate Systems

Cartesian Coordinates

Cartesian coordinates, attributed to René Descartes, employ two perpendicular axes to define the position of a point in a plane. The horizontal axis is typically labelled x, and the vertical axis is labelled y. The position of a point P is given by an ordered pair of numbers (x, y), where x is the distance from the y-axis (abscissa), and y is the distance from the x-axis (ordinate).

Key Concepts

  • Origin: The point where the x and y axes intersect, denoted as O(0, 0).
  • Quadrants: The xy-plane is divided into four quadrants, numbered I to IV, starting from the top right and moving anti-clockwise.
  • Distance Formula: The distance d between two points P1(x1, y1) and P2(x2, y2) is given by d = sqrt((x2 - x1)2 + (y2 - y1)2).

Detailed Definition

Cartesian coordinates, also known as rectangular coordinates, are a type of coordinate system where two or three perpendicular axes define a point in a plane or space. The two-dimensional Cartesian coordinates, denoted as the x- and y-axes, are linear and mutually perpendicular. The x-axis is often considered the horizontal axis, while the y-axis is the vertical axis. The coordinates x and y can range from -∞ to ∞, and an ordered pair (x, y) in two-dimensional Cartesian coordinates is often referred to as a point or a 2-vector.

Example 1: Finding Distance Between Two Points

Consider points A(3, 4) and B(7, 1). Find the distance between A and B.

Using the distance formula: AB = sqrt((7 - 3)2 + (1 - 4)2) AB = sqrt(42 + (-3)2) AB = sqrt(16 + 9) AB = sqrt(25) AB = 5

Thus, the distance between points A and B is 5 units.

Polar Coordinates

Polar coordinates represent a point in a plane by denoting its distance from a reference point (the origin) and the angle from a reference direction. A point P in polar coordinates is represented as P(r, θ), where r is the radial distance from the origin, and θ is the angle from the positive x-axis.

Key Concepts

  • Pole: The fixed point, usually the origin, from which distances are measured.
  • Polar Axis: The ray from the pole in the reference direction.
  • Negative Radius: If r is negative, the point is located |r| units in the direction opposite to θ.
  • Conversion to Cartesian Coordinates: x = r * cos(θ) and y = r * sin(θ).

Example 2: Converting Polar to Cartesian Coordinates

Find the Cartesian coordinates of the point P(5, π/3) in polar coordinates.

Using the conversion formulas: x = 5 * cos(π/3) y = 5 * sin(π/3)

x = 5 * 1/2 y = 5 * sqrt(3)/2

x = 2.5 y = 2.5sqrt(3)

Thus, the Cartesian coordinates of point P are (2.5, 2.5sqrt(3)).

Applications in Geometry

Cartesian Coordinates

  • Line Equations: The equation of a line can be represented as y = mx + c, where m is the slope and c is the y-intercept.
  • Circle Equations: The equation of a circle with centre (h, k) and radius r is (x - h)2 + (y - k)2 = r2.

Polar Coordinates

  • Graphing Curves: Polar coordinates are particularly useful for graphing curves that are symmetrical or involve circles and spirals.
  • Complex Numbers: Polar coordinates can represent complex numbers, where the radial coordinate represents the magnitude and the angular coordinate represents the argument.

Example 3: Equation of a Line

Find the equation of the line passing through points A(2, 3) and B(4, 7).

The slope m of the line AB is given by: m = (y2 - y1) / (x2 - x1) m = (7 - 3) / (4 - 2) m = 2

Using the point-slope form of a line equation y - y1 = m(x - x1), substituting A(2, 3) and m = 2: y - 3 = 2(x - 2) y = 2x - 1

Thus, the equation of the line passing through A and B is y = 2x - 1.

FAQ

Yes, a point can have multiple representations in polar coordinates. This is because the angle θ can be adjusted by adding or subtracting multiples of 2π (or 360 degrees) without changing the position of the point. Additionally, if the radial coordinate r is negative, it means the point is in the opposite direction of the angle θ, which also results in a different representation of the same point. For example, the point P(3, π/4) can also be represented as P(3, 9π/4), P(-3, 5π/4), etc. This flexibility can be advantageous in certain calculations or integrations in calculus.

The angle in polar coordinates, often denoted as theta (θ), indicates the direction of the point from the pole (origin) in relation to the polar axis. It is measured counter-clockwise from the polar axis (which is aligned with the positive x-axis in Cartesian coordinates). The angle can be expressed in degrees or radians, with one full revolution equal to 360 degrees or 2π radians. The angle provides a way to describe the position of a point in a plane, especially when dealing with problems involving symmetry, circles, or spirals, where polar coordinates can offer a simpler or more intuitive representation than Cartesian coordinates.

The equation of a circle in Cartesian coordinates can be derived from the definition of a circle as the set of all points that are equidistant from a fixed point (the centre). If we take a point P(x, y) on the circle and C(h, k) as the centre, the distance between P and C is the radius, r, of the circle. Using the distance formula, we have: sqrt((x - h)2 + (y - k)2) = r. Squaring both sides to eliminate the square root, we obtain the standard form of the equation of a circle: (x - h)2 + (y - k)2 = r2, where (h, k) are the coordinates of the centre and r is the radius.

In Cartesian coordinates, the sign of the x and y coordinates determines the quadrant in which a point lies. Quadrant I: x > 0, y > 0; Quadrant II: x < 0, y > 0; Quadrant III: x < 0, y < 0; Quadrant IV: x > 0, y < 0. In polar coordinates, the angle θ determines the quadrant. Quadrant I: 0 < θ < π/2; Quadrant II: π/2 < θ < π; Quadrant III: π < θ < 3π/2; Quadrant IV: 3π/2 < θ < 2π. The relationship between Cartesian and polar coordinates can be particularly insightful when dealing with problems involving symmetry or rotation, as it allows for an understanding of the position and movement of points in various contexts.

The area of a triangle formed by three points A(x1, y1), B(x2, y2), and C(x3, y3) in the Cartesian coordinate system can be found using the determinant method or Shoelace Theorem. The formula is given by: Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. This formula essentially uses the concept of determinants to calculate the area enclosed by the three points. It's particularly useful as it can handle all types of triangles without needing the height or base to be specifically identified. The absolute value ensures that the area is non-negative, and the 0.5 factor accounts for the fact that we are calculating a triangle's area.

Practice Questions

Consider the points A(2, 3) and B(6, 7) in the Cartesian coordinate system. Find the equation of the line perpendicular to AB that passes through the midpoint of AB.

The midpoint M of a line segment AB with endpoints A(x1, y1) and B(x2, y2) can be found using the midpoint formula: M((x1 + x2)/2, (y1 + y2)/2). Substituting A(2, 3) and B(6, 7) into the formula, we get M(4, 5). The slope of AB, m_AB, can be found using the slope formula: m = (y2 - y1) / (x2 - x1). Substituting A and B, we get m_AB = (7 - 3) / (6 - 2) = 1. The slope of the line perpendicular to AB, m_M, is the negative reciprocal of m_AB, so m_M = -1. Using the point-slope form of the equation of a line, y - y1 = m(x - x1), and substituting M(4, 5) and m_M = -1, we get the equation of the line as y - 5 = -1(x - 4), which simplifies to y = -x + 9.

A point P is represented in polar coordinates as P(5, π/6). Convert the point P into Cartesian coordinates and find the distance from P to the origin O.

To convert point P(5, π/6) from polar to Cartesian coordinates, we use the conversion formulas: x = r * cos(θ) and y = r * sin(θ). Substituting r = 5 and θ = π/6 into the formulas, we get x = 5 * cos(π/6) = 5 * (√3/2) = 5√3/2 and y = 5 * sin(π/6) = 5 * 1/2 = 5/2. Therefore, P in Cartesian coordinates is P(5√3/2, 5/2). To find the distance d from P to the origin O(0, 0), we use the distance formula: d = sqrt((x2 - x1)2 + (y2 - y1)2). Substituting O and P into the formula, we get d = sqrt((5√3/2 - 0)2 + (5/2 - 0)2) = sqrt((25*3/4) + (25/4)) = sqrt(100/4) = sqrt(25) = 5. So, the distance from P to O is 5 units.

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