What is the locus of points equidistant from the lines ax + bc + c = 0?

The given equation of a line can be rewritten in the standard form of a line, which is y = mx + b, where m is the slope and b is the y-intercept. To rewrite the equation, we can solve for y: ax + bc + c = 0 ax + bc = -c y = -(a/b)x - (c/b) The locus of points equidistant from the lines y = -(a/b)x - (c/b) is a line perpendicular to the given lines, with the distance from the origin equal to the absolute value of c/b. This is because the perpendicular bisector of a line segment is equidistant from the endpoints of the segment, and in this case the "endpoints" are the given lines. Since the slope of the given lines is -(a/b), the slope of the perpendicular bisector is b/a. The perpendicular bisector passes through the origin (0,0), so we can find its equation by using the point-slope form of a line: y - 0 = (b/a)x y = (b/a)x This is equivalent to the equation y = mx, where m = b/a, which is the equation of a line passing through the origin with slope b/a. Therefore, the answer is: a line ax - ay + q = 0.