A-maths!!!

Let (x,y) be the coordinates of point P.

d1
= ⊥ distance from point P to L1
= │(x + y + 1)/√[(1)² + (1)²]│
= │(x + y + 1)/√2│

d2
= ⊥ distance from point P to L2
= │(x - y + 2)/√[(1)² + (-1)²]│
= │(x - y + 2)/√2│


a. Since d1 = d2

∴│(x + y + 1)/√2│=│(x - y + 2)/√2│
(x + y + 1) = (x - y + 2) or (x + y + 1) = -(x - y + 2)
2y = 1 or 2x = -3
y = 1/2 or x = -3/2

The equation of locus of point P are:
y = 1/2 or x = -3/2



b. Since d1 = 2d2

∴│(x + y + 1)/√2│=2│(x - y + 2)/√2│
(x + y + 1) = 2(x - y + 2) or (x + y + 1) = -2(x - y + 2)
x - 3y + 3 = 0 or 3x - y + 5 = 0

The equation of locus of point P are:
x - 3y + 3 = 0 or 3x - y + 5 = 0