What is the locus of points equidistant from th... - JAMB Mathematics 2006 Question

Locus of point equidistant from a given straight is the perpendicular bisector of the straight line

∴Gradient of the line ax + by + c = 0

implies by = -ax - c

y = -^a/_bx - c

the gradient (m) = ^a/_b

∴Gradient of the perpendicular (m) = ^b/_a

If P(x,y) is the point intersection of the two lines equation of the perpendicular becomes

y - y₁ = m(x - x₁)

y - y₁ = ^b/_a(x - x₁)

ay - ay₁ = bx - bx₁

ay - bx + bx₁ - ay₁ = 0

ay - bx + b(x₁ - y₁) = 0

implies ay - bx + q