Locus – Problems
1) Find the equation of locus of a point which is a distance 5 from A (4, -3)
Solution: Let p (x, y) be a point in the locus
Given,
A (4, -3)
CP = 5
(x – 4)² + (y + 3)² = 25
x² – 8x + 16 + y² + 6y + 9 – 25 = 0
x² + y² – 8x + 6y = 0
2) Find the equation of locus a point which is equidistance from the point A (-3, 2) and B (0, 4)
Solution: Given points are A (-3, 2), B (0, 4)
let P (x, y) be any point in the locus
PB = PA
PB² = PA²
(x + 3)² + (y – 2)² = (x – 0)² + (y – 4)²
x² + 6x + 9 + y² – 4y + 4 = x² + y² – 8y + 16
6x + 4y = 3 is the equation of the locus
3) Find the equation of locus of a point which is equidistance from the coordinate axes
Solution: Let P (x, y) be any point in the locus
Let PM-perpendicular distance of P from X-axis = |x|
Let PN-perpendicular distance of P from Y-axis = |y|
Given PM = PN
x² – y² = 0.