Usually a set of points which satisfy a particular geometric condition can be represented in the form of an algebraic equation. For example, if we take a fixed distance r and a fixed point O then the set of all points which are at a constant distance r from the fixed point O forms a circle which can be expressed using an algebraic equation. Here the curve (circle) thus obtained is called the locus of the points which satisfy the above condition. Locus is mathematically defined as follows.
“Locus is a curve or the set of points which satisfy the given geometric condition”.
Example:
If we take two points A and B then the set of all points which are equidistant from A and B is the perpendicular bisector of the line segment AB.
Finding the equation of locus:
Step-1: Take any arbitrary point P(x,y) on the locus.
Step-2: Write the geometric condition to be satisfied by P according the problem.
Step-3: Simplify the above equation according to the information provided to get the corresponding algebraic equation of the locus.
Example-1:
Find the locus of a point which is at a distance of 5 units from A(4,-3).
Solution:
Let P(x,y) be a point on the locus.
The given geometric condition is,
PA=5
√[(x-4)2 +(y+3)2] = 5
Squaring on both sides,
(x-4)2 +(y+3)2=25
x2 +16-8x+y2+9+6y=25
x2+y2-8x+6y=0 is the equation of locus.
Example-2:
Find the locus of a point which is equidistant from the points A(-3,2) and B(0,4).
Solution:
Let P(x,y) be a point on the locus.
The given geometric condition is,
PA=PB
√[(x+3)2 +(y-2)2] = √[(x-0)2 +(y-4)2]
Squaring on both sides,
(x+3)2 +(y-2)2 = (x-0)2 +(y-4)2
x2 +9+6x+y2+4-4y= x2 +y2+16-8y
Simplifying the above equation,
6x + 4y = 3, which is the required equation of locus.
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