2006
Mathematics
67ca00ce0c643c71d77dee1f
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 Answer: A line ax - ay +q = 0
Explanation
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
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 + qTHIS WEEK's
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