Definitions
What is a Locus?
A locus (plural: loci) in geometry refers to a set of all points (usually forming a curve or surface) that adhere to a specific condition or property. This condition often involves distance, but can also relate to angles and other geometric entities, providing a rich and varied landscape for exploration and application in various mathematical problems and real-world scenarios.
Distance Condition
A common instance of loci is the set of points that are at a fixed distance from a given point. This locus is a circle if we're working in two dimensions. For example, the locus of points at a distance of 3 units from point A is a circle centred at A with a radius of 3 units.
Angle Condition
Another example is the locus of points that form a fixed angle with two given lines. This could be the angle bisector which equally divides the angle formed by two intersecting lines.
Equidistant Condition
The locus of points equidistant from two given points, A and B, is the perpendicular bisector of the line segment AB. Every point on this line is equidistant from A and B.
Geometric Interpretations
Locus Formed by Moving Point
Consider a point P moving in a plane such that it always remains at a constant distance from a fixed point O. The path traced by P is a circle with O as the centre and OP as the radius.
Locus Formed by Intersection of Two Loci
Sometimes, a locus can be formed by the intersection of two other loci. For instance, the intersection of the loci of points equidistant from two lines (angle bisectors) gives the incenter of a triangle formed by the lines.
Locus in Real-World Applications
Understanding loci can be pivotal in various real-world scenarios, such as determining the path of a satellite around Earth or predicting the trajectory of a planet around the sun. In robotics, loci can help in programming the movement of a robot arm to ensure it moves efficiently while maintaining a certain distance from an object.
Example Questions
Example 1: Locus of Equidistant Points
Consider two points, A(2, 3) and B(5, 7). Find the equation of the locus of points that are equidistant from A and B.
To find the locus equidistant from points A and B, we need to find the perpendicular bisector of AB. The midpoint, M, of AB can be found using the midpoint formula: M = ((x1 + x2)/2, (y1 + y2)/2). Substituting the coordinates of A and B, we get M = ((2 + 5)/2, (3 + 7)/2) = (3.5, 5).
The slope of AB, m1, is (y2 - y1)/(x2 - x1) = (7 - 3)/(5 - 2) = 4/3. The slope of the perpendicular bisector, m2, is -1/m1 = -3/4 because perpendicular lines have slopes that multiply to -1.
Using the point-slope form of a line equation, y - y1 = m(x - x1), and substituting M(3.5, 5) and m2 = -3/4, we get the equation of the locus as y - 5 = -3/4(x - 3.5). Simplifying, we get 4y - 20 = -3x + 10.5, or 3x + 4y = 30.5.
Example 2: Locus Formed by Moving Point
A point P is moving such that it is always equidistant from the points C(1, 2) and D(4, 6). Find the equation of the locus of P.
Since P is equidistant from C and D, the locus of P is the perpendicular bisector of CD. Using similar steps as in Example 1, we find the midpoint, E, of CD: E = ((1 + 4)/2, (2 + 6)/2) = (2.5, 4).
The slope of CD, m1, is (6 - 2)/(4 - 1) = 4/3. The slope of the perpendicular bisector, m2, is -1/m1 = -3/4.
Using the point-slope form of a line equation, y - y1 = m(x - x1), and substituting E(2.5, 4) and m2 = -3/4, we get the equation of the locus as y - 4 = -3/4(x - 2.5). Simplifying, we get 4y - 16 = -3x + 7.5, or 3x + 4y = 23.5.
FAQ
Yes, loci can be represented in three-dimensional space, extending the 2D concepts by introducing a third dimension. In 3D, loci can represent a set of points that satisfy certain conditions in space, not just on a plane. For example, the locus of points equidistant from a fixed point in 3D is a sphere. Similarly, the locus of points equidistant from a fixed line in 3D is a cylindrical surface. Understanding loci in 3D involves considering spatial relationships and can be applied in fields like physics, engineering, and computer graphics to solve problems involving spatial configurations and relationships.
The concept of loci is foundational to understanding conic sections like ellipses and hyperbolas. An ellipse is the locus of points where the sum of the distances from two fixed points (foci) is constant. In contrast, a hyperbola is the locus of points where the absolute difference of the distances from two fixed points (foci) is constant. These definitions inherently involve the concept of loci as they describe a set of points that satisfy particular geometric conditions relative to fixed points. Understanding loci provides insights into the properties and applications of these conic sections in various mathematical and real-world contexts.
The perpendicular bisector in the concept of loci is significant because it represents the set of all points that are equidistant from two given points. It is derived by first finding the midpoint of the line segment connecting the two points, and then determining the line that is perpendicular to the segment and passes through the midpoint. Mathematically, if the two points are A(x1, y1) and B(x2, y2), the midpoint, M, is given by M((x1 + x2)/2, (y1 + y2)/2). The slope of AB, m1, is (y2 - y1)/(x2 - x1), and the slope of the perpendicular bisector, m2, is -1/m1. The equation of the perpendicular bisector can then be found using the point-slope form of a line equation and substituting M and m2.
Loci play a crucial role in engineering and architecture by helping to determine the path or position of an object, point, or area under specific conditions. For instance, in structural engineering, loci can be used to determine the path of a moving part in machinery or to find the optimal location for a component that needs to be equidistant from multiple points. In architecture, loci can be used to design curves and surfaces that are equidistant from a set of points, ensuring structural integrity and aesthetic appeal. Understanding the geometric properties of loci allows professionals to create designs and structures that adhere to specific spatial constraints and requirements.
When a locus is subjected to transformations like translations or rotations, the entire set of points moves or rotates according to the transformation rules, maintaining its geometric properties. For instance, if a locus is defined as a circle with a certain radius around a point, translating the center of the circle to a new location will translate the entire locus (circle) to a new position while maintaining its shape and size. Similarly, if a locus is defined by a set of conditions relative to a line, rotating the line will rotate the entire locus. Understanding how loci interact with transformations allows for the exploration of their properties and applications under various conditions and operations in the coordinate plane.
Practice Questions
To find the locus equidistant from points A and B, we need to find the perpendicular bisector of AB. The midpoint, M, of AB can be found using the midpoint formula: M = ((x1 + x2)/2, (y1 + y2)/2). Substituting the coordinates of A and B, we get M = ((3 + 7)/2, (4 + 8)/2) = (5, 6).
The slope of AB, m1, is (y2 - y1)/(x2 - x1) = (8 - 4)/(7 - 3) = 1. The slope of the perpendicular bisector, m2, is -1/m1 = -1 because perpendicular lines have slopes that multiply to -1.
Using the point-slope form of a line equation, y - y1 = m(x - x1), and substituting M(5, 6) and m2 = -1, we get the equation of the locus as y - 6 = -1(x - 5). Simplifying, we get y = -x + 11.
To find the locus equidistant from a point and a line, we need to find the parabola with focus R and directrix PQ. The directrix, PQ, can be represented by the line equation y - y1 = m(x - x1), where m is the slope of PQ and (x1, y1) is a point on PQ. The slope, m, is (y2 - y1)/(x2 - x1) = (7 - 3)/(6 - 2) = 1. Using P(2, 3) in the line equation, we get y - 3 = 1(x - 2), or y = x + 1 as the equation of the directrix.
The parabola with focus R(4, 5) and directrix y = x + 1 has the equation (x - h)^2 = 4p(y - k), where (h, k) is the vertex and p is the distance from the vertex to the focus or directrix. The vertex is the midpoint between the focus and its reflection across the directrix. To find this reflection, note that the line perpendicular to the directrix and passing through R is y = -x + 9. The intersection of y = x + 1 and y = -x + 9 is found by setting the y’s equal: x + 1 = -x + 9, or x = 4. Substituting x = 4 into y = x + 1 gives y = 5, so the reflection is S(4, 5), and the vertex is the midpoint of RS, which is (4, 5). Since R is above the directrix, p is positive and equal to the distance from R to the directrix, which is the absolute value of the y-coordinate of R minus the y-coordinate of a point on the directrix with the same x-coordinate, or |5 - (4 + 1)| = 0. Thus, the equation of the parabola is (x - 4)2 = 0(y - 5), or x = 4.
