Geometry and Locus: Definition, Theorem, Examples- Embibe
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  • Written By Keerthi Kulkarni
  • Last Modified 20-05-2022
  • Written By Keerthi Kulkarni
  • Last Modified 20-05-2022

Geometry and Locus: Loci of Geometrical Figures, Shapes

Geometry and Locus: Geometric shapes and entities are defined in modern Mathematics as a collection of points that satisfy a given criterion. The word locus is derived from the word location. A locus is a set of points in geometry that satisfy a condition or scenario for a shape or figure. The plural form of locus is loci.

There are different types of geometrical figures, such as parallel lines, perpendicular bisectors, angle bisectors, circles, ellipses, parabolas and hyperbolas drawn with the help of locus which undergoes certain conditions. Let us learn more about locus and its theorems in this article.

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Geometry and Locus

Geometry in Mathematics deals with the shapes, angles, dimensions, and sizes of a wide range of objects encountered in daily life. There are two-dimensional and three-dimensional shapes in Euclidean geometry. Coordinate geometry explains how points, lines, and planes are used in basic geometry. Points, lines, angles and curves are some of the geometrical figures.

What is a Locus?

We are all aware that the Earth travels in an elliptical orbit around the Sun. Observe that the Earth’s positions at different times are combined to generate an elliptical orbit. In this situation, the locus is the arc that connects all of the Earth’s positions. As a result, the produced locus is the geometrical figure, ellipse.

In Mathematics, a locus is a curve or shape or surface produced by all the points satisfying a given equation of the coordinate relation. The locus defines all shapes, including circles, ellipses, parabolas, and hyperbolas, as a set of points satisfying a specific condition.

A locus is a set of all points whose position is determined by one or more defined conditions in geometry.

How to Determine a Locus?

The steps to determine the locus are listed below:

Step 1: Observe the condition given
Step 2: Draw the points of the locus
Step 3: Connect all the points of the locus
Step 4: Identify the pattern.
Step 5: Describe the locus.

Locus of Points

Let us determine the locus of a point, which is equidistant from two points, \(X\) and \(Y\). It is known as a straight line. In the below figures, the locus is a straight line that is equidistant from the points \(X\) and \(Y\). It is constructed by drawing the arcs of the same radius.

Locus Theorems

The set of points with certain conditions is called locus. Depending on the different conditions, numerous geometrical figures such as parallel lines, perpendicular bisectors, pairs of parallel lines and angle bisectors are formed.

  • Locus Theorem 1
    A pair of parallel lines are formed as the locus of points that are equidistant \((d)\) from the line \(m\) on either side.
  • Locus Theorem 2
    The perpendicular bisector is the locus of points that are equidistant from the ends of the line segment \(AB\)
  • Locus Theorem 3
    The line that lies halfway or midway between two parallel lines is the locus of points that equidistant from the two parallel lines \(m_1\) and \(m_2\)
  • Locus Theorem 4
    The locus of all the points at the same distance from both sides of an angle can be determined using this. The angle bisector is the locus. An angle bisector is the locus equidistant from the sides of the angle and present on the interior of an angle.
  • Locus Theorem 5
    The locus is a pair of lines that bisects the angle formed by the two intersecting lines, say \(m_1\) and \(m_2\) This aids in the identification of the region created by all points that are at the same distance from two crossing lines. A pair of lines should bisect the angle formed to build the locus.

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Locus as Other Geometric Figures

  • Circle
    The locus of all the points at a certain distance from a fixed point is a circle. Here, the ‘centre’ of the circle is the fixed point. The fixed distance is known as the circle’s ‘radius’.
  • Concentric Circles
    The circle concentric with the specified circles and placed halfway between them is the locus of points equidistant from two concentric circles.

Geometry and Locus

If the sum of the lengths from \(P\) to \(F_1\) and to \(F_2\) is a constant, given two points \(F_1\) and \(F_2\) (the foci), then the locus of points formed is the ellipse. The locus is a hyperbola, where the absolute value of the difference between the distances from \(P\) to \(F_1\) and to \(F_2\) is a constant.

The locus of points in a cassini oval is such that the product of the distances from \(P\) to \(F_1\) and to \(F_2\) is a constant. A parabola is a set of points where the distance between two points \(F_1\) and \(F_2\) (the focus) equals the distance between two lines (the directrix).

  • Parabola
    A parabola is a set of points or locus that is equidistant from a fixed point and a line. The focus is the fixed point, and the line is the parabola’s directrix.
  • Ellipse
    Ellipse is a set of points that satisfy the requirement that the sum of two foci point distances is constant.
  • Hyperbola
    Two focal points are equidistant from the centre of the semi-major axis of a hyperbola. The collection of points that satisfy the requirement that the absolute value of the difference between the distances to two specified foci is a constant is known as a hyperbola.
  • Oval Cassini
    The locus of points in a cassini oval is such that the product of the distances from and to \(P\) is a constant.

Equation of Locus

Locus is the path traced out by a moving point under one or more given conditions. Every locus (curve) has a mathematical equation known as the locus equation. There is no formula for determining the locus.

The steps to finding locus of points in two-dimensional geometry are as follows:

Step 1: Assume that the moving point’s coordinates are \((x_1,\,y_1)\).
Step 2: Apply the geometrical criteria to \((x_1,\,y_1)\), resulting in a relationship between \(x_1\) and \(y_1\).
Step 3: In the resulting equation, replace \(x_1\) with \(x\) and \(y_1\) with \(y\).
Step 4: The equation obtained is the locus equation.

Standard equations of some of the geometrical figures are listed below:

Geometrical FigureEquation of Locus
Line\(y = mx + C\)
Parallel lines\(x = k\) or \(y = k\)
Circle\((x – h)^2 + (y – k)^2 = r^2\)
Parabola\(y^2 = 4ax\)
Ellipse\(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\)
Hyperbola\(\frac {x^2}{a^2} – \frac {y^2}{b^2} = 1\)

Solved Examples – Geometry and Locus

Q.1. What is the equation for the locus of points equidistant between \(A(-2,\,3)\) and \(B(6,\,-5)\)?
Solution:
Let the point be \(P(x,\,y)\),
Given, point \(P(x,\,y)\) is equidistant from \(A(-2,\,3)\) and \(B(6,\,-5)\).
So, \(PA = PB\)
\(\Rightarrow {\left( {x + 2} \right)^2} + {\left( {y – 3} \right)^2} = {\left( {x – 6} \right)^2} + {\left( {y + 5} \right)^2}\)
\(\Rightarrow {\left( {x + 2} \right)^2} – {\left( {x – 6} \right)^2} = {\left( {y + 5} \right)^2} – {\left( {y – 3} \right)^2}\)
\(\Rightarrow \left( {2x – 4} \right)\left( 8 \right) = \left( {2y + 2} \right)\left( 8 \right)\)
\(\Rightarrow 2x – 4 = 2y + 2\)
\(\Rightarrow x – 2 = y + 1\)
\(\therefore x = y + 3\) or \(x – y = 3\)

Q.2. Determine the equation for a moving point’s locus that is always equidistant from the points \((2,\,-1)\) and \((3,\,2)\). The locus represents which geometric figure?
Solution:
Let the point be \(P(x,\,y)\) and \(A(2,\,-1)\) and \(B(3,\,2)\).
Given, point \(P(x,\,y)\) is equidistant from \(A(2,\,-1)\) and \(B(3,\,2)\).
So, \(PA = PB\)
\(\Rightarrow PA^2 = PB^2\)
\(\Rightarrow {\left( {x\; – \;2} \right)^2}\; + \;{\left( {y\; + \;1} \right)^2}\; = \;{\left( {x\; – \;3} \right)^2}\; + \;{\left( {y\; – \;2} \right)^2}\)
\(\Rightarrow {x^2}\; – \;4x\; + \;4\; + \;{y^2} + \;2y\; + \;1 = \;{x^2}\;–\;6x\; + \;9\; + \;{y^2}\;–\;4y\; + \;4\)
\(\Rightarrow 2x + 6y = 8\)
\(\therefore \,x + 3y = 4\)
Clearly, the equation is a first-degree equation in \(x\) and \(y\).
Hence, the locus of the point is a straight line whose equation is \(x + 3y = 4\).

Q.3. The sum of a moving point’s distances between \((c,\,0)\)  and \((-c,\,0)\) is always \(2a\) units. Find the equation for the moving point’s locus.
Solution:
Let \(P(h,\,k)\) be the moving point.
Consider \(A(c,\,0)\) and \(B(-c,\,0)\)
According to the question,
\(PA = PB = 2a\)
\(PA = 2a – PB\)
\(\Rightarrow P{A^2}\; = \;4{a^2} + P{B^2} – 4a \cdot PB\)
\(\Rightarrow P{A^2} – P{B^2} = 4{a^2} – 4a \cdot PB\)
\(\Rightarrow \left[ {{{\left( {h\; – \;c} \right)}^2} + {{\left( {k\; – \;0} \right)}^2}} \right] – \left[ {{{\left( {h\; + \;c} \right)}^2} + {{\left( {k\; – \;0} \right)}^2}} \right] = 4{a^2} – 4a \cdot PB\)
\(\Rightarrow – 4hc = 4{a^2} – 4a \cdot PB\)
\(\Rightarrow a \cdot PB = {a^2} + hc\)
\(\Rightarrow {a^2}\left[ {{{\left( {h + c} \right)}^2} + {{\left( {k – 0} \right)}^2}} \right] = {\left( {{a^2} + hc} \right)^2}\)
\(\Rightarrow {a^2}\;\left[ {{h^2} + {c^2}\; + \;2hc\; + {k^2}} \right]\; = \;{a^4}\; + \;2{a^2}hc\; + \;{h^2}{c^2}\)
\(\Rightarrow {a^2}{h^2} – {h^2}{c^2} + {a^2}{k^2} = {a^4} – {a^2}{c^2}\)
\(\Rightarrow \left( {{a^2} – {c^2}} \right){h^2} + {a^2}{k^2} = {a^2}\left( {{a^2} – {c^2}} \right)\)
\(\therefore \,\frac{{{h^2}}}{{{a^2}}} + \frac{{{k^2}}}{{{a^2} – {c^2}}} = 1\)
Hence, the required equation to the locus of the point \(P\) is \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{a^2} – {c^2}}} = 1\).

Q.4. Help the Priya find which geometric figure is formed by the locus of a point equidistant from the \(y-\) axis and the point \((-1,\,2)\).
Solution:
Let the locus of point \(P(x,\,y)\)
According to the question, distance from \(P\) to \((-1,\,2) =\) Distance from \(P\) to \(y-\) axis.
\(\sqrt {{{\left( {x – \left( { – 1} \right)} \right)}^2} + {{\left( {y + 2} \right)}^2}} = x\)
\( \Rightarrow {\left( {x + 1} \right)^2} + {\left( {y – 2} \right)^2} = {x^2}\)
\(\Rightarrow {x^2} + 2x + 1 + {y^2} – 4y + 4 = {x^2}\)
\( \Rightarrow {y^2} – 4y + 5 + 2x = 0\)
\(\Rightarrow {y^2} – 4y + 4 = – 2x – 1\)
\(\therefore \,{\left( {y – 2} \right)^2} = – 1\left( {2x + 1} \right)\)
This equation is in the form of \({\left( {y – k} \right)^2} = a\left( {x – h} \right)\)
Hence, it is the locus of a parabola.

Q.5. Draw the locus of points which is equidistant from the line segment.
Solution:
Let us consider a line segment \(AB\) as shown below
The locus of the set of points that are equidistant from the given line segment is shown below:

Summary

Geometry is the figure formed by the locus of points, which are bounded with some predefined conditions. For example, the earth orbit around the sun is the locus. There are different types of geometrical figures formed by loci such as straight line, parallel lines, perpendicular bisector, angular bisector etc.

Depends on the various conditions such as the distance of the point from the focus such as ellipse, oval casinni, parabola and hyperbola. Circle is the locus of points which are at a fixed distance from fixed point.

By taking any random point on the required locus and considering the required conditions, we can construct a locus. There are different types of loci equations associated with the different geometrical figures.

Frequently Asked Questions (FAQs)

Below are the frequently asked questions on Geometry and Locus:

Q.1. What is a locus in geometry?
Ans: A locus is a set of points that satisfy a specific condition or situation for a shape or figure in geometry. The plural form of locus is loci. The term locus comes from the term location.

Q.2. What are the examples of locus?
Ans: The orbit of earth around the sun is the locus, which is in the shape of ellipse. The sphere is a three-dimensional figure in which the set of points that are equidistant from a given point.

Q.3. How do you find the loci of a point?
Ans: The steps to finding the locus of points in two-dimensional geometry are as follows:
• Assume that any random point \(P(x,\,y)\) on the locus exists.
• Create an equation based on the provided conditions.
• To get the locus equation, simplify it.

Q.4. What is the locus of a circle.
Ans: A circle’s locus is a set of points on a plane that are all the same distance from the centre point.

Q.5. What is the locus of \(y = mx + c\)?
Ans: The locus of the equation \(y = mx + c\) is a straight line.

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