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LOCUS
Definition: The set of all points (and only those points) which satisfy the given geometrical
condition(s) (or properties) is called a locus
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The set of points in a plane which are at a constant distance r from a given point C is a locus
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2
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Here the locus is a straight line and it is the perpendicular bisector of the line segment joining A
and B
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An equation of a locus is an algebraic description of the locus
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(iii) Apply the proper formula of coordinate geometry and translate the geometric
condition(s) into an algebraic equation
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The equation thus obtained is the required equation of locus
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1
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Sol
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Given A(4, –3)
P(x,y)

5
A(4,–3)

Given that CP = 5
⇒ CP2 = 25
⇒ (x – 4)2 + (y + 3)2 = 25
⇒ x2 – 8x + 16 + y2 + 6y + 9 – 25 = 0
∴ Equation of the locus of P is:
x2 + y2 – 8x + 6y = 0
2
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Sol
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3
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Sol
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Find the equation of locus of a point which is equidistant from the coordinate axes
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Let P(x, y) be any point in the locus
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= |x|
Let PN = perpendicular distance of P from Y-axis
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Find the equation of locus of a point equidistant from A(2, 0) and the Y-axis
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Given point is A (2, 0)
Y
P(x,y)
Let P(x, y) be any point in the locus
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= |x|
X
Given that is PA = PN
O
A(2,0)
2
2
⇒ PA = PN
⇒ (x – 2)2 + (y – 0)2 = x2
⇒ x2 – 4x + 4 + y2 = x2
∴ Locus of P is y2 – 4x + 4 = 0
6
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Y

P(x,y)
y

O

Sol
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Now OP2 = x2 + y2

M

X

Given condition is OP2 = 4y ⇒ x2 + y2 = 4y
Equation of the locus of P is x2 + y2 – 4y= 0
Find the equation of locus of a point P such that PA2 + PB2 = 2c2, where A= (a, 0),
B(–a, 0) and 0 < |a| < |c|
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Let P(x, y) be a point in locus
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II
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Find the equation of locus of P, if the line segment joining (2, 3) and (–1, 5) subtends a
right angle at P
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Given points A (2, 3), B (–1, 5)
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90°
Given condition is: ∠APB = 90°
7
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The ends of the hypotenuse of a right angled triangle are (0, 6) and (6, 0)
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Sol
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3
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Sol
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Find the equation of locus of P, if A(4, 0), B(–4, 0) and |PA – PB| = 4
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Same as above
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Find the equation of locus of a point, the sum of whose distances from (0, 2) and
(0, –2) is 6
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Given points are A (0, 2) and B (0, –2)
Let P(x, y) be any point in the locus
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45
45
5
9
6
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Sol
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⇒

7
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Find the equation of locus of P, so that the
area of triangle PAB is 9
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Given points are A(5, 3), B(3, –2)
Let P(x, y) be a point in the locus
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1 x −5 y−3
⇒
=9
2 3 − 5 −2 − 3

⇒

x −5 y−3
−2

−5

= 18

⇒| −5x + 25 + 2y − 6 |= 18

⇒| −5x + 2y + 19 |= 18
⇒ −5x + 2y + 19 = ±18
⇒ −5x + 2y + 19 = 18 or − 5x + 2y + 19 = 18
⇒ 5x − 2y − 1 = 0 or 5x − 2y − 37 = 0

∴ Locus of P is :
(5x – 2y – 1)(5x – 2y – 37) = 0
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Find the equation of locus of a point which forms a triangle of area 2 with the point
A(1, 1) and B(–2, 3)
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Same as above
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(2x + 3y – 1)(2x + 3y – 9) = 0
9
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Sol
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Given points are A(2, 3), B(2, –3)
Given that PA : PB = 2 : 3
⇒ 3PA = 2PB
⇒ 9PA2 = 4PB2
⇒ 9[(x–2)2 + (y–3)2] = 4[(x–2)2 + (y+3)2]
⇒ 9[x2 – 4x + 4 + y2 – 6y + 9] = 4[x2 – 4x + 4 + y2 + 6y + 9]
⇒ 5x2 + 5y2 – 20x – 78y + 65 = 0 which is the equation of locus
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A(1, 2), B(2, –3) and C(–2, 3) are three points
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Show that the equation to the locus of P is 7x – 7y + 4 = 0
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Let P(x, y) be a point in locus
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A straight rod of length 9 slides with its ends A, B always on the X and Y-axes
respectively
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Sol
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Let OA = a and OB = b and AB =9
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B(0,b)
Let G (x,y ) be the centroid of ∆OAB
a b
G
But Coordinates of G of ∆OAB are  , 
3 3
a b
Therefore,  ,  = (x, y)
3 3
a
b
⇒ = x, = y ⇒ a = 3x, b = 3y
3
3
2
But OA + OB2 = AB2 and given AB = 9
⇒ a2 + b2 = 81
⇒ 9(x2 + y2) = 81
∴ Equation to the locus of P is x2 + y2 = 9
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Find the equation of the locus of a point which is at a distance 5 from (–2, 3) in a plane
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x2 + y2 + 4x – 6y – 12 = 0
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Find the equation of locus of a point P, if the distance of P from A(3, 0) is twice the
distance of P from B(–3, 0)
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x2 + y2 + 10x + 9 = 0
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Find the locus of the third vertex of a right angled triangle, the ends of whose
hypotenuse are (4, 0) and (0, 4)
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x2 + y2 – 4x – 4y = 0
4
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Ans
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5
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Find the equation of locus of P so that the
area of the triangle PAB is 8
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Ans

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