What comes to mind when you hear the word locus? A locus is a set of points that satisfy a specific condition or situation for a shape or figure. The concept of locus is very important in the study of various geometrical shapes. The plural of locus is loci.

A circle is the locus of points in the plane that are equidistant from a particular location. Furthermore, a sphere is the set of points in 3D space that is equidistant from a particular location. The locus of the points also describes other shapes like an ellipse, parabola, and hyperbola. In this article, we will discuss about Locus in detail. Scroll down to learn more about Locus of Points!

Definition of Locus

In geometry, a locus is a set of points that satisfy a specific condition or situation for a shape or figure. The locus is pluralized as loci. The term locus comes from the term location. Before the twentieth century, geometric shapes were once thought of as an entity or a location where points might be located or moved. However, in contemporary mathematics, entities are defined as a collection of points that satisfy a certain condition.

A locus is a curve or shape formed by all the points satisfying a specific equation of the relationship between the coordinates or by a point, line, or moving surface in mathematics. The locus defines all shapes as a set of points, including circles, ellipses, parabolas, and hyperbolas. You’ve probably heard the term ‘location’ in real life. The term locus is the root of the word location. The term locus refers to the location of something. The locus describes when an object is located somewhere or when something occurs at a certain location.

Other helpful links on the locus concept:

Locus Equation of Parabola	Locus Equation of Ellipse
Locus Equation of Hyperbola	Alternate Locus in Circle

A locus is a collection of points that satisfy some condition (typically forming a curve or surface). A circle is the locus of points in the plane that are equidistant from a given location, while a sphere is the set of points in a three-dimensional space that are equidistant from a given point.

See below how the locus of a point equidistant from two given circles is created with a dotted line? The fact is that its locus is still a circle.

What is Locus of Points?

In two-dimensional geometry, the locus of points is a curve or a line. A shape is defined in geometry by the locus of points. A circle is the locus of all points equidistant from a fixed point known as its centre. The locus of the points also defines other shapes like an ellipse, parabola, and hyperbola.

Example: The locus of a point whose sum of distances from the two fixed points is constant will be an ellipse.

The locus of all points whose distance is constant from a fixed point will be a circle. Similarly, the locus of all points that are equidistant from a line segment is basically the perpendicular bisector of that line segment.

Examples on Locus

There are some examples of locus which are helpful in geometry are as follows:

1. The locus of a point whose distance from a fixed point \(A\) is constant, say \(r\), is a circle with \(A\) as the centre and \(r\) as the radius.

2. A pair of parallel lines lying on either side of the line \(l\) at a distance \(d\) from the line \(l\) is considered the locus of a point at a fixed distance \(d\) from the line \(l\).

3. Perpendicular bisectors of a line segment are the locus of the points which are equidistant from the endpoints of the line segment. This example aids in the identification of the region formed by all points that are at the same distance from point \(A\) and point \(B\). The perpendicular bisector of the line segment \(AB\) should be created.

4. The bisector of an angle is the locus that is present on the interior of an angle and is equidistant from the sides of the angle.

5. The locus of points equidistant from line segment \(AB\) is shown below

6. The locus of points that bisect the angle formed by the two intersecting lines say \(l_1\) and \(l_2\) will also be a pair of straight lines but these lines are perpendicular to each other.

7. The shapes like parabola, ellipse, and hyperbola, all come under the concept of locus. The locus of a point whose distance from a fixed point and a fixed-line is the same is basically a parabola, and the locus of a point whose sum of distances from two fixed points is constant will be an ellipse. Similarly, the locus of a point whose differences from two fixed points is constant will be a hyperbola.

8. A locus (plural loci) in genetics is a specific, fixed location on a chromosome where a gene or genetic marker is found. Each chromosome has numerous genes, each of which is located at a distinct locus.

Solved Examples on Locus

Q.1. Find the locus of a point that moves at a constant distance of two units above the \(x\)-axis.
Ans: Two units from \(x\)-axis. In geometry, a locus is a set of all points whose location satisfies or is determined by one or more specified conditions.
The equation of a line which is \(2\) units above \(x\)−axis for all points is \(y=2.\)

Q.2. Find the value of \(k\) and \(b\), if the points \(P(-3,1)\) and \(Q(2,b)\) lie on the locus of \(x^2-5x+ky=0\)
Ans: Given that \(P(-3,1)\), \(Q(2,b)\)
Point \(P(-3,1)\) lies on the locus \(x^2-5x+ky=0\)……..(1) \( \Rightarrow {\left({ – 3} \right)^2} – 5\left({ – 3} \right) + k\left(1 \right) = 0\)
\(\Rightarrow k = \, – 24\)
Let us substitute \(?(2,?)\) on the locus, as the point lies on locus \(x^2-5x+ky=0\) and value of \(k\)
\( \Rightarrow 4 – 10 + \left({ – 24} \right)b = 0\)
\(\therefore \,b = – \frac{1}{4}\)
Hence \({\rm{k = }}\,{\rm{ – 24}}\), \(b = \frac{{ – 1}}{4}.\)

Q.3. If \(Q\) is a point on the locus of \(x^2+y^2+4x-3y+7=0\), then find the equation of locus of \(P\) which divides segment \(OQ\) externally in the ratio \(3:4\), where \(O\) is the origin.
Ans: Let \(Q\) be \(\left({x,\,y}\right)\) & \(P\left({h,\,k} \right)\)
We know that the coordinate of point \(P\) which divides the other two point externally, is given by
\(P\left({h,\,k} \right) = \left({\frac{{m{x_2} – n{x_1}}}{{m – n}},\frac{{m{y_2} – n{y_1}}}{{m – n}}} \right)\) ……..(1)
Given, \(m=3\), \(n=4\); \(O({x_1}, {y_1})=(0, 0), Q({x_2}, {y_2})=(x, y)\)
\(\therefore \,P = \left({\frac{{3x – 4(0)}}{{3 – 4}},\frac{{3y – 4(0)}}{{3 – 4}}} \right)\)
\(\left({h,\,k} \right) = \left({\frac{{3x}}{{ – 1}},\,\frac{{3y}}{{ – 1}}} \right)\)
\(\therefore \,h = – 3x\); \(k=-3y\)
\(x = \frac{ { – h}}{3}\); \(y = \frac{ { – k}}{3}\)
\(\therefore \,Q\left({x,\,y} \right) = Q\left({ – \frac{h}{3},\,\frac{k}{3}} \right)\)
As \(?\) is a point on the locus \( {x^2} + {y^2} + 4x – 3y + 7 = 0\)
Substitute \(Q\left({ – \frac{h}{3}, – \frac{k}{3}} \right) {\text{}}\) in \( {x^2} + {y^2} + 4x – 3y + 7 = 0\)
\( \Rightarrow {\left( { – \frac{h}{3}} \right)^2} + {\left( { – \frac{k}{3}} \right)^2} + 4\left( { – \frac{h}{3}} \right) – 3\left( { – \frac{k}{3}} \right) + 7 = 0\)
\( \Rightarrow \frac{{{h^2}}}{9} + \frac{{{k^2}}}{9} – \frac{{4h}}{3} + k + 7 = 0\)
\( \Rightarrow {h^2} + {k^2} – 12h + 9k + 63 = 0\)
Substitute \(h=x\),\(k=y\)
\(\therefore {x^2} + {y^2} – 12x + 9y + 63 = 0\)
Hence, the equation of locus \(P\) is \( {x^2} + {y^2} – 12x + 9y + 63 = 0\)

Q.4. The sum of the distance of a moving point from the points \((4,0)\) and \((-4,0)\) is always 10 units. Find the equation of the locus of the moving point.
Ans:
Let \(P(h,k)\) be any point and \((4,0)\),\((-4,0)\) are the points.Let \(A(4,0)\) & \(B(-4,0)\)
Given \(PA+PB=10\) …. (1)
Using the distance between two point formula \(PA = \sqrt {{{\left({{x_2} – {x_1}} \right)}^2} + {{\left({{y_2} – {y_1}} \right)}^2}}.\)
\(P(h,k)\),\(A(4,0)\)
\(PA = \sqrt {{{\left({{x_2} – {x_1}} \right)}^2} + {{\left({{y_2} – {y_1}} \right)}^2}}.\)
\( = \sqrt { { {(h – 4)}^2} + { {(k – 0)}^2}} \)………..(2)
Now \(P(h,k)\),\(B(-4,0)\)
\(PB = \sqrt {{{\left({{x_2} – {x_1}} \right)}^2} + {{\left({{y_2} – {y_1}} \right)}^2}} \)
\( = \sqrt { { {(h + 4)}^2} + { {(k – 0)}^2}} \)……..(3)
Put equation (2) & (3) in equation (1)
\( \Rightarrow \sqrt {{{(h – 4)}^2} + {k^2}} + \sqrt {{{(h + 4)}^2} + {k^2}} = 10\)
\( \Rightarrow \sqrt {{{(h – 4)}^2} + {k^2}} = 10 – \sqrt {{{(h + 4)}^2} + {k^2}} \)
Squaring on both sides
\( \Rightarrow {(h – 4)^2} + {k^2} = {\left({10 – \sqrt {{{(h + 4)}^2} + {k^2}} } \right)^2}\)
\( \Rightarrow {(h – 4)^2} + {k^2} = 100 + {(h + 4)^2} + {k^2} – 2 \times 10 \times \sqrt {{{(h + 4)}^2} + {k^2}} \)
\( \Rightarrow {h^2} + 16 – 8h = 100 + {h^2} + 16 + 8h – 20\sqrt {{{(h + 4)}^2} + {k^2}} \)
\( \Rightarrow – 16h – 100 = – 20\sqrt {{{(h + 4)}^2} + {k^2}} \)
\( \Rightarrow – 4(4h + 25) = – 20\sqrt {{{(h + 4)}^2} + {k^2}} \)
\( \Rightarrow 4h + 25 = 5\sqrt {{{(h + 4)}^2} + {k^2}} \)
Squaring on both sides
\({(4h + 25)^2} = {\left({5\sqrt {{{(h + 4)}^2} + {k^2}} } \right)^2}\)
\( \Rightarrow 16{h^2} + 625 + 200h = 25\left({{{(h + 4)}^2} + {k^2}} \right)\)
\( \Rightarrow 16{h^2} + 625 + 200h = 25{h^2} + 200h + 400 + 25{k^2}\)
\( \Rightarrow 9 {h^2} + 25 {k^2} – 225 = 0\)
Divide throughout by \(225\)
\( \Rightarrow \frac{{9{h^2}}}{{225}} + \frac{{25{k^2}}}{{225}} = 1\)
\( \Rightarrow \frac{{{h^2}}}{{25}} + \frac{{{k^2}}}{9} = 1\)
Substitute \(ℎ=?\),\(?=?\)
\(\therefore \frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1\)
Hence, the locus of the moving point is \(\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1\)

Q.5. If the locus of point \(P(9\,\cos \alpha ,9\,\sin \alpha )\) for all values of \(α\) is of the form \(x^2+y^2=a^2\). Then, what is the value of \(a^2\)?
Ans: Let \(P(h,k)\) be any point on the required path
\(\therefore h = 9\cos \alpha ;k = 9\sin \alpha \)
\( \Rightarrow \frac{h}{9} = \cos \alpha \) ……(1)
\( \Rightarrow \frac{k}{9} = \sin \alpha \) ……(2)
From equation (1) & (2)
We know that, \( {\cos ^2}\theta + {\sin ^2}\theta = 1\)
\(\therefore {\cos ^2}\alpha + {\sin ^2}\alpha = 1\)
\( \Rightarrow {\left({\frac{h}{9}} \right)^2} + {\left({\frac{k}{9}} \right)^2} = 1\)
\( \Rightarrow \frac{{{h^2}}}{{81}} + \frac{{{k^2}}}{{81}} = 1\)
\( \Rightarrow \frac{{{h^2} + {k^2}}}{{81}} = 1\)
\( {h^2} + {k^2} = 81\)
Replace \(ℎ=?\),\(?=?\)
Hence, locus of \(P=x^2+y^2=81\) ………..(3)
Comparing the equation (3) with the given equation \(x^2+y^2=a^2\), we get
\(a^2=81\)
Hence, the required value of \(?^2\) is \(81\).

Summary

In this article, we have learnt about the Locus, Locus of points and its example and got to know that it plays an essential role in the mathematical world. In mathematics, it is a curve or other shape formed by all points satisfying a specific equation of the relation between the coordinate or by a point, line, or moving surface. The locus defines all shapes as a set of points, including circles, ellipses, parabolas, and hyperbolas.

FAQs on Locus

Q.1. What is a locus of a point?
Ans: In two-dimensional geometry, the locus of points is a curve or a line. A shape is defined in geometry by the locus of points. Assume that a circle is the centre of all equidistant locations. The locus of the points also defines other shapes like an ellipse, parabola, and hyperbola.

Q.2. What is a chromosome locus?
Ans: A locus is a location or “address” on a chromosome where all members of a species have the same gene for a certain feature. It can also refer to where a mutation or other genetic marker is found. Any pair of homologous chromosomes can contain a particular locus.

Q.3. What is a locus in genetics?
Ans: A locus (plural loci) in genetics is a specific, fixed location on a chromosome where a gene or genetic marker is found. Each chromosome has numerous genes, each of which is located at a distinct locus.

Q.4. What is the locus of a circle?
Ans: A locus is a collection of points that satisfy a set of criteria. The locus of a circle is a set of points that are always equidistant from the centre of the circle.

Q.5. What is meant by locus in maths?
Ans: In maths, the locus is a set of points that satisfy some condition (typically forming a curve or surface).

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