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 \(q = bx_{1} - ay_{1}\)
\(\therefore \text{The perpendicular bisector of the line is } ay - bx + q = 0\)
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