Question

Show that the path of a moving point, such that its distance from two lines `3x - 2y = 5` and `3x + 2y = 5` are equal, is a straight line?

Answer 1

see explanation.

Explanation:

Recall that the distance from a point `(h,k)` to the line `ax+by+c=0` is given by :
`d=|ah+bk+c|/sqrt(a^2+b^2)`,
given lines are :
`3x-2y-5=0, and 3x+2y-5=0`
let `P(h,k)` be any point which is equidistant to the two given lines,
by the given conditions,
`=> |(3h-2k-5)|/sqrt(3^2+(-2)^2)= |(3h+2k-5)|/sqrt(3^2+2^2)`
`=> |(3h-2k-5)|= |(3h+2k-5)|`
`=> 3h-2k-5=+-(3h+2k-5)`
case 1 : `3h-2k-5=3h+2k-5`
`=> k=0`, which is a straight line.
case 2 : `3h-2k-5=-(3h+2k-5)`
`=> 6h=10, => h=5/3`, which is also a straight line.
Hence, the path of the moving points is a straight line, as shown in the figure.

Footnotes : Angle bisector of two lines is the locus of a point which is equidistant from the two lines, which means an angle bisector has equal perpendicular distance from the two lines.