Locus: Definition, Theorems and Examples

In this article, we will cover the concepts of Locus and its equation. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of Eleven questions have been asked on JEE MAINS( 2013 to 2023) from this topic which include one in 2015, one in 2017, two in 2018, two in 2019, two in 2021, two in 2022, one in 2023.

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This Story also Contains

1. What is Locus?
2. Steps to Finding the Equation of Locus
3. Theorems of Locus
4. Locus of Points
5. Locus of a Circle
6. Solved Examples Based on Loci and its Equation.

What is Locus?

When a point moves in a plane under certain geometrical conditions, then the point traces out a path, This path of the moving point is known as the locus of this point. The curve described by a variable point on the plane that moves under a given condition or a set of conditions is called its locus.

For example, let point $\mathrm{O}(0,0)$ be a fixed point (i.e. origin) and a variable point $P(x,y)$ is in the same plane. If point P moves in such a way that the distance OP is constant ${\mathbf{r}}_{}$, then point P traces out a circle whose centre is $O(0,0)$ and radius is ${\mathbf{r}}_{}$.

Since $\mathrm{OP}=\mathrm{r}$, we have $(x-a{)}^2+(y-b{)}^2=r^2$ which is the equation of locus of point P.

Steps to Finding the Equation of Locus

1. Consider the point $(h,k)$ whose locus is to be found.

2. Express the given condition as an equation in terms of the known quantities and unknown parameters.

3. Eliminate the parameters so that the resultant equation consists only of locus coordinates $\text{ h, k, }$ and known quantities.

4. Now, replace the locus coordinate $(h,k)$ with $(x,y)$ in the resultant equation.

Illustration

Find the equation of the locus of the point which is at a constant distance of 5 units from a point $(2,3)$

Solution

Let $\mathrm{A}=(2,3)$ and $\mathrm{B}=(\mathrm{h},\mathrm{k})$
$\mathrm{AB}=$ constant $=5$
$(\mathrm{AB}{)}^2=5^2=25$
$(\mathrm{AB}{)}^2=(\mathrm{h}-2{)}^2+(\mathrm{k}-3{)}^2=25$
$h^2-4h+4+k^2-6k+9=25$

The equation of locus is

$x^2+y^2-4x-6y-12=0$

Theorems of Locus

Theorem 1: Locus of a point that moves in a plane such that its distance from a fixed point in the same plane is always constant. This is a locus of the circle. The fixed point is called the center of the circle and the constant distance is the radius of the circle.

Theorem 2: The locus of a point that moves in a plane such that the ratio of its distance from a fixed point to its perpendicular distance from a fixed straight line is constant, is called conic section or conic. This fixed point is called the focus of the conic and the fixed line is called the directrix of the conic.

Theorem 3: The locus at a fixed distance “d” from the line “m” is considered as a pair of parallel lines that are located on either side of “m” at a distance “d” from the line “m”.

Theorem 4: The locus which is equidistant from the two given points say A and B, are considered as perpendicular bisectors of the line segment that joins the two points.

Theorem 5: The locus which is equidistant from the two parallel lines say m₁ and m₂, is considered to be a line parallel to both the lines m₁ and m₂ and it should be halfway between them.

Theorem 6: The locus which is present on the interior of an angle equidistant from the sides of an angle is considered to be the bisector of the angle.

Theorem 7: The locus which is equidistant from the two intersecting lines say m₁ and m₂, is considered to be a pair of lines that bisects the angle produced by the two lines m₁ and m₂.

Locus of Points

The locus of points defines a shape in geometry. Suppose, a circle is the locus of all the points which are equidistant from the centre. Similarly, the other shapes such as an ellipse, parabola, hyperbola, etc. are defined by the locus of the points.

The locus is defined only for curved shapes. These shapes can be regular or irregular. Locus is not described for the shapes having vertex or angles inside them.

Locus of a Circle

The circle is defined as the set of all points equidistant from a fixed point, where the fixed point is the centre of the circle and the distance of the sets of points from the centre is the radius of the circle.

Recommended Video Based on Locus and its Equations

Solved Examples Based on Loci and its Equation.

Example 1: If the point $(\alpha ,\frac{7\sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x\cos \theta +y\sin \theta =7,\theta \in (0,\frac{\pi }{2})$ between the co-ordinates axes, then $\alpha$ is equal to [JEE MAINS 2023]

Solution:

$(\alpha ,\frac{7\sqrt{3}}{3})$

$\begin{array}{rl}& x\cos \theta +y\sin \theta =7\\ & x-\text{ intercept }=\frac{7}{\cos \theta }\\ & y-\text{ intercept }=\frac{7}{\sin \theta }\\ & A:(\frac{7}{\cos \theta },0)B:(0,\frac{7}{\sin \theta })\end{array}$

Locus of mid pt M : (h, k)
$\begin{array}{rl}& \mathrm{h}=\frac{7}{2\cos \theta },\mathrm{k}=\frac{7}{2\sin \theta }\\ & \frac{7}{2\sin \theta }=\frac{7\sqrt{3}}{3}\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{3}\\ & \alpha =\frac{7}{2\cos \theta }=7\end{array}$

Hence, the answer is 7.

Example 2: A point P moves so that the sum of squares of its distances from the points $(1,2)$ and $(-2,1)$ is $14$. Let $f(x,y)=0$ be the locus of p which intersects the x-axis at the points A, B and the y-axis at the points $\mathrm{C},\mathrm{D}$. Then the area of the quadrilateral $\text{ ACBD }$ is equal to [JEE MAINS 2022]

Solution: Let the coordinates of point P be $(x,y)$.

$\begin{array}{rl}& \Rightarrow (\mathrm{x}-1{)}^2+(\mathrm{y}-2{)}^2+(\mathrm{x}+2{)}^2+(\mathrm{y}-1{)}^2=14\\ & \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}-3\mathrm{y}-2=0\end{array}$

Put $x=0$

$\begin{array}{rl}& \Rightarrow {\mathrm{y}}^2-3\mathrm{y}-2=0\\ & \Rightarrow \mathrm{y}=\frac{3\pm \sqrt{17}}{2}\\ & \Rightarrow {\mathrm{x}}^2+\mathrm{x}-2=0\\ & \Rightarrow (x+2)(x-1)=0\\ & \therefore \mathrm{A}(-2,0),\mathrm{B}(1,0),\mathrm{C}(0,\frac{3+\sqrt{17}}{2}),\mathrm{D}(0,\frac{3-\sqrt{17}}{2})\\ & \text{ Area }=\frac{1}{2}\times 3\times \sqrt{17}\\ & =\frac{3\sqrt{17}}{2}\end{array}$

Hence, the answer is $\frac{3\sqrt{17}}{2}$.

Example 3: The locus of a point, which moves such that the sum of squares of its distances from the points $(0,0),(1,0),(0,1),(1,1)$ is 18 units, is a circle of diameter $\text{ d }$. Then $d^2$ is equal to ____ [JEE MAINS 2021]

Solution: Let the point be $P(h,k)$

As per the question

$\begin{array}{rl}& 2^2+k^2+(h-1{)}^2+k^2+h^2+(k-1{)}^2+(h-1{)}^2+(k-1{)}^2=18\\ & 4h^2+4k^2-4h-4k=14\\ & \Rightarrow x^2+y^2-x-y-\frac{7}{2}=0\end{array}$

Diameter $d=2\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{7}{2}}$
$\begin{array}{rl}& =2\sqrt{\frac{8}{2}}\\ & =4\end{array}$
Hence, $d^2=16$.

Hence, the answer is 16.

Example 4: Let $A$ be a fixed point $(0,6)$ and $B$ be a moving point $(2t,0)$. Let $M$ be the mid-point of $AB$ and the perpendicular bisector of $AB$ meets the $y$-axis at $C$. The locus of the mid-point $P$ of $MC$ is [JEE MAINS 2021]

Solution

The coordinates of mid-point M of AB will be $(\mathrm{t},3)$.
slope of AB :
$m_{AB}=\frac{0-6}{2t-0}=\frac{-3}{t}$
slope of MC :
$m_{MC}=\frac{t}{3}[AsAB\perp MC]$
Equation of MC$y-3=\frac{t}{3}(x-t)$
coordinates of $C:x=0,y=3-\frac{t^2}{3}$
so the Mid-point of MC will be$\begin{array}{rl}& x=\frac{0+t}{2}=\frac{t}{2},y=\frac{3+3-\frac{t^2}{3}}{2}=\frac{6-\frac{t^2}{3}}{2}\cdot 3-\frac{t^2}{6}\\ & \Rightarrow t=2x\Rightarrow y=3-\frac{(2x{)}^2}{6}\\ & \Rightarrow 3y=9-2x^2\\ & \Rightarrow 2x^2+3y-9=0\end{array}$

Hence, the answer is $2x^2+3y-9=0$.

Example 5: A straight line through a fixed point $(2,3)$ intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is: [JEE MAINS 2018]

Solution: Intercept form of a straight line -

$\frac{x}{a}+\frac{y}{b}=1$

$a$ and $b$are the $x$-intercept and $y$-intercept respectively.

The equation of PQ is

$\frac{x}{h}+\frac{y}{k}=1$

i.e put $(x,y)$ $\Rightarrow (2,3)$

$\begin{array}{rl}& \frac{2}{h}+\frac{3}{k}=1\\ & 3x+2y=xy\end{array}$

Hence, the answer is $3x+2y=xy$.