What is the locus of points equidistant from th... - JAMB Mathematics 2006 Question
What is the locus of points equidistant from the lines ax + bc + c = 0?
A
A line bx - ay +q = 0
B
A line ax - ay +q = 0
C
A line bx + ay +q = 0
D
A line bx + ay +q = 0
correct option: b
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
∴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
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