Class 9 Opt Maths: Locus

Equation of Locus

A locus is a set of points which satisfy certain geometric conditions. Many geometric shapes are most naturally and easily described as loci. For example, a circle is the set of points in a plane which are a fixed distance  from a given point the center of the circle.

Problems involving describing a certain locus can often be solved by explicitly finding equations for the coordinates of the points in the locus. Here is a step-by-step procedure for finding plane loci:

Step 1: If possible, choose a coordinate system that will make computations and equations as simple as possible. 
Step 2: Write the given conditions in a mathematical form involving the coordinates  and . 
Step 3: Simplify the resulting equations. 
Step 4: Identify the shape cut out by the equations.

Step 1 is often the most important part of the process since an appropriate choice of coordinates can simplify the work in steps 2-4 immensely.

Locus

To every point in a plane there corresponds on ordered pair of real numbers and to every ordered pair of real numbers, there corresponds a point on a plane. If we have an algebraic relation between x and y, we get a set of points in the plane for different values of x and y satisfying the relation. If we move a set of points determined by some geometrical condition, we can represent it by an algebraic relation. This algebraic relation is called equation and the set of points in the plane is called Locus.

The locus of a point is the path traced out by the point moving under given geometrical condition( or conditions). Alternatively, the locus is the set of all those points, which satisfy the given geometrical conditions (or conditions).

For example:

1. Let a point P move such that its distance from a fixed line is always equal to d. The point P will trace out a straight line CD parallel to the fixed line. Thus, the locus of the moving point P is the straight line.
2. Consider the set of all points on the x-axis. This is the locus of a point whose ordinate is zero.
3. Let a point P move in a plane such that its distance from a fixed point, says B, is always equal to r. The point P will trace out a circle with centre B and radius r. Thus, the locus of the moving point is the circle.
4. Consider the locus of a point whose distance from the origin is 5 units. If the coordinates of any points satisfying the condition be (x,y), then (x-0)2+(y-0)2=52 is the equation of the locus.Hence, the equation of the locus of a point whose distance from the origin is equal to 5 units is x2+y2=25.The locus obviously is a circle with centre at the origin and radius 5 units.

Remarks

1. Every point which satisfies the given geometrical conditions (or conditions) lies on the locus.
2. A point which does not satisfy the given geometrical condition ( or conditions ) cannot lie on the locus.
3. Every point which lies on the locus satisfies the given geometrical condition (or conditions).
4. A point which does not lie on the locus cannot satisfy the given geometrical condition (or conditions).
5. The locus of a point moving in a plane under a given geometrical conditions is always a straight or a curved line.
6. To find the locus of moving points, plot some points satisfying the given geometrical condition, and then join these points.

To find the equation of a set of points (locus) with given geometrical conditions, the working rule is

1. Suppose (x, y) be any points in the set,
2. Find a relation between x and y satisfying the given conditions.
3. The relation is the required equation of the locus.

Notes

1. If any point belongs to the Locus, then the point satisfies the equation of the locus.
2. Any point satisfying the equation of Locus must lie on the Locus.
3. Any point out the locus does not satisfy the equation of the locus.

Methods of Finding the Equation of Locus

1. Take P(x,y) as a moving point in the coordinate plane. Represent the information in a suitable figure.
2. Find the distances according to the information provided.
3. Simplify the expressions by using algebraic operations. Make sure that there is no common number in all the terms. Also, make sure that first term is not negative. Final result or equation so obtained is the equation of the locus.
   
   
   

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Some problems:

1. Define locus.
Ans: A locus is a set of points which satisfy certain geometric conditions.

Many geometric shapes are most naturally and easily described as loci.

For example, a circle is the set of points in a plane which are a fixed distance r from a given point  O the center of the circle.

2. Find the equation of the locus of the point P, if the points A (1, 2), B (3, 4) are at an equal distance from P?
Solution: Let P(x , y) be any point.
Distance of this point from (1,2)is √{ (x-1)² + (y-2)²
Distance of this point from (3,4 ) is √{ (x-3)² + (y-4)².
As  the given condition ,
√{ (x-1)² + (y-2)² } = √{ (x-3)² + (y-4)²}
Squaring we get,
      (x-1)² + (y-2)² =  (x-3)² + (y-4)².
⇒  x² - 2x + 1 + y² - 4y + 9 = x² - 6x + 9 + y² - 8y + 16
⇒  4x + 4y  = 20
 ∴   x + y = 5 is the equation of  locus.

3. A point moving in such a manner that three times of distance from the x-axis is grater by 7 than 4 times of its distance form the y-axis; find the equation of its locus.
Solution: Let P (x, y) be any position of the moving point on its locus.
Then the distance of P from the x-axis is y and its distance from the y-axis is x.
i.e., 3y – 4x = 7,
Which is the required equation to the locus of the moving point.

4. Find the equation to the locus of a moving point which is always equidistant from the points (2, -1) and (3, 2).
Solution: Let A (2, -1) and B (3, 2) be the given points and (x, y) be the co-ordinates of a point P on the required locus.
Then,
       PA² = (x - 2)² + (y + 1)²
and  PB² = (x - 3)² + (y - 2)² .
By problem,
      PA = PB
⇒   PA² = PB²
⇒  (x - 2)² + (y + 1)² = (x - 3)² + (y - 2)²
⇒   x² - 4x + 4 + y² +2y+1 = x² – 6x + 9 + y² –4y+4
⇒   2x + 6y = 8
⇒  x + 3y = 4 .
Which is the required equation to the locus of the moving point.

5. A and B are two given point whose co-ordinates are (-5, 3) and (2, 4) respectively. A point P moves in such a manner that PA : PB = 3 : 2.

Find the equation to the locus traced out by P.

Solution: Let P(x, y) be the co-ordinates of any position of the moving point on its locus.
Given,
     PA : PB = 3:2
⇒  3  PB = 2  PA
⇒  9  PB² = 4  PA²
⇒  9[(x - 2)² + (y - 4)²] = 4[(x + 5)² + (y - 3)²]
⇒  9 [x² - 4x + 4 +y² - 8y + 16] = 4[x² + 10x + 25 + y²- 6y + 9]
⇒   5x² + 5y² – 76x – 48y + 44 = 0 .
Therefore , the required equation to the locus traces out by P is 5x² + 5y² – 76x – 48y + 44 = 0 .

6. Find the locus of a moving point which forms a triangle of area 21 square units with the point (2, -7) and (-4, 3).
Solution: Let the given point be A (2, -7) and B (-4, 3) and the moving point P (say), which forms a triangle of area 21 square units with A and B, have co-ordinates (x, y).

Thus, by question area of the triangle PAB is 21 square units.
Hence, we have 
      ½ | (6 – 4y - 7x) – ( 28 + 3x + 2y) | = 21
⇒ |6 – 28 - 4y – 2y - 7x – 3x | = 42
⇒ 10x + 6y + 22 = ±42

Either, 10x + 6y + 22 = 42     i.e., 5x + 3y = 10
Or 10x + 6y + 22 = - 42        i.e., 5x + 3y + 32 = 0.

Therefore, the required equation to the locus of the moving point is 5x + 3y = 10 or, 5x + 3y + 21 = 0.

7. The sum of the distance of a moving point from the points (c,0) and (-c, 0) is always 2a units. Find the equation to the locus of the moving point.
Solution: Let P be the moving point and the given points be A (c,0) and B (-c, 0).
If (x, y) be the co-ordinates of any position of P on its locus then by question,
     PA + PB = 2a
⇒ PA = 2a - PB
⇒ PA² = 4a² + PB² – 4a ∙ PB
⇒ PA² – PB² = 4a² – 4a ∙ PB
⇒ [(x - c)² +(y - 0)²] - [(x + c)² +(y - 0)²] = 4a² – 4a. PB
⇒ -4xc = 4a² – 4a∙PB
⇒  a ∙ PB = a² + hc
⇒  a² ∙ PB² = (a² + xc)² (squaring both sides)
⇒  a² [(x + c)² + (y - 0)²] = (a² + xc)²
⇒  a² [x² + c² + 2xc + y²] = a⁴ + 2a²xc + x²c²
 ⇒  x²/a² + y²/(a² – c²) = 1

Therefore, the required equation to the locus of P is x²/a² + y²/(a² – c²) = 1.

8. Find the equation of the locus of a moving point P(x, y) which is always at a distance of 5 units from a fixed point Q (2, 4).
Solution: Let r be the distance from fixed point and (2,4) = ( x₁, y₁) .
Then,
     PQ = r
⇒ (x – x₁)²+ (y – y₁)² = r² 
⇒ (x – 2)² + (y – 4)² = 5²
⇒ x² – 4x + 4 + y² – 8y + 16 = 25
⇒  x² + y²– 4x – 8y – 5 = 0 is the required equation.

9. A (2, 0) and B (0, -2) are two fixed points. Point P moves with a ratio so that AP: PB = 1: 2.  Find the equation of the locus of point P.
Solution: Let, co-ordinate of P be (x, y).

According to question,
     AP:PB=1:2
⇒ 2AP = PB
⇒ 2√[(x−2)²+(y−0)²]=√[(x−0)²+(y−(−2))²]
Square both sides to eliminate the square roots we get
    4[(x−2)²+y2]=x²+ (y+2)²
⇒ 4(x²−4x+4+y²)= x²+y²+4y+4
⇒ 4x²−16x+16+4y²=x²+y²+4y+4
⇒  3x²+3y²−16x−4y+12=0.
Hence, the equation of the locus of point P is 3x²+3y²−16x−4y+12=0.

10. Find the value of k so that the point (4,5) lies in the locus x² + y² + kx - 8y - 25 = 0.
Solution: Here,
The point (4,5) lies in the locus x² + y² + kx - 8y - 25 = 0.
So, put x  = 4  & y = 5, we get
    4² + 5² + k.4 - 8.5 - 25 = 0.
⇒  16 + 25 + 4k - 40 - 25 = 0
⇒   4k = 24
⇒  k = 6

Thus, k = 4.
 
11. (4, 4) is a point on the locus whose equation is y² = kx.  Show that (16,8) is another point on the locus.
Solution: Here,
      (4, 4) is a point on the locus whose equation is y² = kx. 
So, 4² =4 . k
i.e  k = 4.

Hence, equation of locus is  
     y² = 4x.

Again put (x,y) = (16,8) in this locus, we get
     8² = 4⨯ 16
i.e. 64 = 64 (True)

Thus, (16,8) is another point on the given locus.

12. A(3,0) & B(-3,0) are two fixd points. Find the locus of a point P at which AB substend a right angle at P.
Solution: Let P(x,y) be the point.
According to question, AB substend a right angle at P.
So,
     AP² + BP² = AB² 
⇒   (x - 3)² + ( y - 0)² + (x + 3)² + ( y - 0)²= (3 + 3)² + (0 - 0)²
⇒  x² - 6x +9 + y² + x² + 6x + 9 + y² = 6²
⇒  2x² + 2y² = 36 - 18
⇒  x² + y² = 9 is the required equation of locus.
 
13. A(a,0) & B(-a,0) are two fixd points. Find the locus of a point P at which AB substend a right angle at P.
Solution: Let P(x,y) be the point.

According to question, AB substend a right angle at P.
So,
    AP² + BP² = AB² 
⇒   (x - a)² + ( y - 0)² + (x + a)² + ( y - 0)²= (a + a)² + (0 - 0)²
⇒   x² - 2ax +a² + y² + x² + 2ax + a² + y² = (2a)²
⇒   2x² + 2y² = 4a² - 2a²
⇒   x² + y² = a²  is the required equation of locus.
 
14. For all value of the co-ordinates of a moving point P are (a cos θ, b sin θ); find the equation to the locus of P.
Solution: Let (x, y) be the co-ordinates of any point on the locus traced out by the moving point P.
Then we shall have ,
      x = a cos θ
⇒   
xa = cos θ .................(i)   and
      y = b sin θ
 ⇒    xa = sin θ .................(ii)

Squaring and adding these two equations,
     x²a² + y²b² = cos²θ + sin²θ
⇒    x²a² + y²b² = 1, which is the required equation to the locus of P.

15. Find the locus of the point which is at a constant distance of 10 units from (2,-8) .
Solution: P(x,y)be the point in Locus.
If A = (2, - 8) , given geometrical condition is
      PA = 10
⇒  PA² = 100
⇒   (x -2 )² + ( y+8)² = 100
⇒  x² - 4x + 4 + y² + 16y + 64 = 100
⇒  x² + y² - 4x +16y - 32 = 0.
Thus, equation of locus is x² + y² - 4x +16y - 32 = 0.

16. Find the equation to the set of points equidistant from (-1,-1)and (4,2).
Solution:  Let A =( -1,-1) and B =( 4,2)

If P( x,y ) be the point , then the geometrical condition satisfied by P is

      PA = PB

⇒ PA² = PB²

⇒  ( x +1) ² + ( y + 1 )² = ( x - 4 )² + (y - 2) ²

⇒ x² +2x +1 +y² +2y +1 = x² - 8x + 16 + y² - 4y + 4

⇒ 5x + 3y = 9.

Therefore, the equation of locus is 5x + 3y = 9.

17. Find the locus of a point which moves in such a way that its distance from (3,0) is always equal to its distance from the origin.
Solution: Let A = (3,0) and O = (0,0).

Let P( x,y ) be a point satisfying the geometrical condition.,

     PA = OP
⇒ PA² = OP²
⇒ ( x - 3 )² + y² = x² +y²
⇒ 6x = 9
⇒ 2x = 3.

Therefore, the equation of locus of P is 2x - 3 = 0.

18. Find the locus of a point that is equidistant from the point (1, 3) and (2, 4).
Solution: Let P(x, y) be any point on the locus. It is known that the distance of P from A(1, 3) and B(2, 4) is always same

i.e. PA = PB.

Now,

PA = √{(x-1)²+(y-3)²}

PB = √{(x-2)²+(y-4)²}

Thus,

PA² = PB²

⇒ (x-1)² + (y-3)² = (x-2)² + (y-4)²
⇒ x² – 2x + 1 + y² –6y+9 = x² – 4x + 4 + y² –8y+16
⇒ 2x + 2y = 10

Hence, x + y = 5, which is the required equation.

19. Find the equation of the locus of a point which moves in such a way that its distance from (-1, 4) is always 9.
Solution: Let P(x, y) be a point moving in such a way that its distance from (-1, 4) is always 9 .

i.e. PA = 9.

Therefore,
     (PA)² = 81
⇒ (x+1)² + (y - 4)² = 81
⇒ x² + 2x + 1 + y² – 8y + 16 = 81.

Thus, x² + y² + 2x – 8y – 64 = 0, which is the required equation.

20. Find the locus of the point which is at a constant distance of 10 units from (2,-8) .
Solution: P(x,y)be the point in Locus.
If A = (2, - 8) , given geometrical condition is
      PA = 10
⇒ PA² = 100
⇒  (x -2 )² + ( y+8)² = 100
⇒ x² - 4x + 4 + y² + 16y + 64 = 100
⇒  x² + y² - 4x +16y - 32 = 0.
Thus, equation of locus is x² + y² - 4x + 16y - 32 = 0.