**Definition:**

Generally locus of the point can be defined as the path traced by a point when it is moving with a specific geometrical conditions.

In the diagram below the point A(x₁,y₁) which lies on the circle whose distance from the point O(h,k) is a constant "r". The fixed point O is called the center of the circle and the distance from O to A is called radius "r".

**Example 1:**

Find the locus of the point which is equidistant from (-1,1) and (4,-2).

**Solution:**

Let A (-1,1) and B(4,-2) be the given points and P(x,y) be the any point on locus.Since it is equidistant from the locus point we have the condition AP = BP

Distance of AP = √(x₂ - x₁)² + (y₂ - y₁)²

A (-1,1) P (x,y)

x₁ = -1 , y₁= 1 ,x₂ = x and y₂ = y

= √(x-(-1))² + (y-1)²

= √(x+1)² + (y-1)²

= √x² + 1² + 2(x) (1) + y² + 1² - 2(y) (1)

= √x² + 1² + 2x + y² + 1² - 2y

= √x² + y² + 2x - 2y + 2 ------------ (1)

Distance of BP = √(x₂ - x₁)² + (y₂ - y₁)²

B(4,-2) P (x,y)

x₁ = 4 , y₁= -2 ,x₂ = x and y₂ = y

= √(x-4)² + (y-(-2))²

= √x² + 4² - 2 (4) (x) + (y + 2) ²

= √x² + 16 - 8x + y² + 2² + 2(y) (2)

= √x² + 16 - 8x + y² + 4 + 4y

= √x² + y² - 8x + 4y + 20 ------------ (2)

(Distance of AP) = (Distance of BP)

√x² + y² + 2x - 2y + 2 = √x² + y² - 8x + 4y + 20

taking squares on both sides

[√x² + y² + 2x - 2y + 2]² = [√x² + y² - 8x + 4y + 20]²

x² + y² + 2x - 2y + 2 = x² + y² - 8x + 4y + 20

x² - x² + y² -y² + 2x +8x -2y - 4y + 2 -20= 0

10 x- 2y -18 = 0

5x - y -9 = 0

So the lo-cus of the point P(x,y) is 5x - y -9 = 0

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