Mathematics

What is the locus of points equidistant from the lines ax + by + c =…

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)