This lesson will be focused on equation to a locus. I’ll again split it into two parts due to its length. Equation to a LocusA formal(ish) definition: “The equation of a curve is the relation which exists between the coordinates of all points on the curve, and which does not hold for any point not on the curve”. Let us try to understand what this means. If I write an equation, say x + y = 4 and tell you that this represents a line which looks like this… … it means that if you take any random point lying on this line, take its x-coordinate and add it to the y-coordinate, you’ll always get 4 as the sum (because the equation says x + y = 4). And if you take any other point not on the line, and add its coordinates together, you’ll never get the sum as 4. (For now, don’t worry about why x + y = 4 should look like a line, and not something different, e.g. a circle. We’ll see that later.) Thus, finding out the equation to a locus means finding out the relation that holds true between the x and y coordinates of all points on the locus. Let’s find out equations to all the loci we covered previously. Example 1 Find the locus of the point moving on a plane which is at a fixed distance 5 units from a the X axis. Let P(x, y) be the moving point. Now, the distance of this point from the X axis is its y-coordinate. Since this distnace is given to be 5, we can write this relation in the form of an equation as y = 5 Well, that’s it! We got the equation representing the locus. Going in the reverse order, the equation y = 5 is the equation of the locus (or curve), every point on which has the y-coordinate as 5, or every point being at a distance of 5 units from the X-axis. There is also another possibility of y = -5, also a line parallel to the X-axis, at a distance of 5 units, but lying below the axis. Example 2 Find the locus of a point which is at a fixed distance 4 from the origin. This locus (or path) was a circle. To find its equation, we’ll start by converting the given condition into a mathematical form, using the formulas we know. Let P(x, y) be the moving point. It is given that OP = 4 (where O is the origin). How can we convert this into mathematical form? OP is the distance between O and P which can be written as \(\sqrt{(x-0)^2+(y-0)^2}\) Hence, we can write \(\sqrt{(x-0)^2+(y-0)^2}\) = 4 ⇒ x2 + y2 = 16 Therefore, the equation to the locus under the given conditions is x2 + y2 = 16. Example 3 Find the locus of a point such that it is equidistant from two fixed points, A(1, 1) and B(2, 4). We had figure out that this was the perpendicular bisector of the line joining the points. Now to the equation. Let P(x, y) be the moving point. According to the condition, PA = PB. This can be written as \(\sqrt{(x-1)^2+(y-1)^2}=\sqrt{(x-2)^2+(y-4)^2}\). After squaring both sides and simplifying, we get the equation as x + 3y = 9 Lesson Summary
That’s it for this part. Questions involving the locus will become a little more complicated as we proceed. The next part will cover the remaining examples. In geometry, a locus (plural: loci) (Latin word for "place", "location") is a set of all points (commonly, a line, a line segment, a curve or a surface), whose location satisfies or is determined by one or more specified conditions.[1][2] In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property. The use of the singular in this formulation is a witness that, until the end of the 19th century, mathematicians did not consider infinite sets. Instead of viewing lines and curves as sets of points, they viewed them as places where a point may be located or may move. Until the beginning of the 20th century, a geometrical shape (for example a curve) was not considered as an infinite set of points; rather, it was considered as an entity on which a point may be located or on which it moves. Thus a circle in the Euclidean plane was defined as the locus of a point that is at a given distance of a fixed point, the center of the circle. In modern mathematics, similar concepts are more frequently reformulated by describing shapes as sets; for instance, one says that the circle is the set of points that are at a given distance from the center.[3] In contrast to the set-theoretic view, the old formulation avoids considering infinite collections, as avoiding the actual infinite was an important philosophical position of earlier mathematicians.[4][5] Once set theory became the universal basis over which the whole mathematics is built,[6] the term of locus became rather old-fashioned.[7] Nevertheless, the word is still widely used, mainly for a concise formulation, for example:
More recently, techniques such as the theory of schemes, and the use of category theory instead of set theory to give a foundation to mathematics, have returned to notions more like the original definition of a locus as an object in itself rather than as a set of points.[5] Examples from plane geometry include:
Other examples of loci appear in various areas of mathematics. For example, in complex dynamics, the Mandelbrot set is a subset of the complex plane that may be characterized as the connectedness locus of a family of polynomial maps. To prove a geometric shape is the correct locus for a given set of conditions, one generally divides the proof into two stages:[10]
(distance PA) = 3.(distance PB) Find the locus of a point P that has a given ratio of distances k = d1/d2 to two given points. In this example k = 3, A(−1, 0) and B(0, 2) are chosen as the fixed points. P(x, y) is a point of the locus ⇔ | P A | = 3 | P B | {\displaystyle \Leftrightarrow |PA|=3|PB|} ⇔ | P A | 2 = 9 | P B | 2 {\displaystyle \Leftrightarrow |PA|^{2}=9|PB|^{2}} ⇔ ( x + 1 ) 2 + ( y − 0 ) 2 = 9 ( x − 0 ) 2 + 9 ( y − 2 ) 2 {\displaystyle \Leftrightarrow (x+1)^{2}+(y-0)^{2}=9(x-0)^{2}+9(y-2)^{2}} ⇔ 8 ( x 2 + y 2 ) − 2 x − 36 y + 35 = 0 {\displaystyle \Leftrightarrow 8(x^{2}+y^{2})-2x-36y+35=0} ⇔ ( x − 1 8 ) 2 + ( y − 9 4 ) 2 = 45 64 . {\displaystyle \Leftrightarrow \left(x-{\frac {1}{8}}\right)^{2}+\left(y-{\frac {9}{4}}\right)^{2}={\frac {45}{64}}.}This equation represents a circle with center (1/8, 9/4) and radius 3 8 5 {\displaystyle {\tfrac {3}{8}}{\sqrt {5}}} . It is the circle of Apollonius defined by these values of k, A, and B. Second exampleLocus of point C A triangle ABC has a fixed side [AB] with length c. Determine the locus of the third vertex C such that the medians from A and C are orthogonal. Choose an orthonormal coordinate system such that A(−c/2, 0), B(c/2, 0). C(x, y) is the variable third vertex. The center of [BC] is M((2x + c)/4, y/2). The median from C has a slope y/x. The median AM has slope 2y/(2x + 3c).
The locus of the vertex C is a circle with center (−3c/4, 0) and radius 3c/4. Third exampleThe intersection point of the associated lines k and l describes the circle A locus can also be defined by two associated curves depending on one common parameter. If the parameter varies, the intersection points of the associated curves describe the locus. In the figure, the points K and L are fixed points on a given line m. The line k is a variable line through K. The line l through L is perpendicular to k. The angle α {\displaystyle \alpha } between k and m is the parameter. k and l are associated lines depending on the common parameter. The variable intersection point S of k and l describes a circle. This circle is the locus of the intersection point of the two associated lines. Fourth exampleA locus of points need not be one-dimensional (as a circle, line, etc.). For example,[1] the locus of the inequality 2x + 3y – 6 < 0 is the portion of the plane that is below the line of equation 2x + 3y – 6 = 0.
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