What is the locus of points equidistant from the lines ax + by + c = 0?
- A.
A line bx – ay +q = 0 - B.
A line ax – by +q = 0 - C.
A line bx + ay +q = 0 - D.
A line ax + by +q = 0
Correct Answer: Option B
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 = frac{-a}{b}x – frac{c}{b})
(Gradient = frac{-a}{b})
(therefore text{The gradient of the perpendicular bisector} = frac{b}{a})
If P(x,y) is the point intersection of the two lines equation of the perpendicular becomes
(y – y_{1} = m(x – x_{1}))
(y – y_{1} = frac{b}{a}(x – x_{1}))
(frac{y – y_{1}}{x – x_{1}} = frac{b}{a})
(ay – ay_{1} = bx – bx_{1})
(ay – bx + bx_{1} – ay_{1} = 0)
Let (q = bx_{1} – ay_{1})
(therefore text{The perpendicular bisector of the line is } ay – bx + q = 0)