The locus of a point equidistant from the intersection of lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a

The locus of a point equidistant from the intersection of two lines is the perpendicular bisector of the line segment joining the intersection point of the lines. Therefore, we need to find the intersection point of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0, and then find the perpendicular bisector passing through that point. To find the intersection point of the two lines, we can solve their simultaneous equations. Multiplying the first equation by 4 and the second equation by 3, we get: 12x - 28y + 28 = 0 12x - 18y + 3 = 0 Subtracting the second equation from the first, we get: -10y + 25 = 0 Solving for y, we get: y = 5/2 Substituting this value of y in either of the original equations, we get: 3x - 7(5/2) + 7 = 0 3x = 17/2 x = 17/6 So, the intersection point of the lines is (17/6, 5/2). Now, we need to find the perpendicular bisector of the line segment joining this point with any other point equidistant from the intersection point. Since the question does not specify any other point, we can assume that we are looking for the perpendicular bisector of the line segment joining the intersection point with itself. This means we need to find the equation of the line passing through the point (17/6, 5/2) and perpendicular to the line joining the intersection point with itself, which is the x-axis. The line passing through (17/6, 5/2) and perpendicular to the x-axis has a slope of 0, so its equation is of the form y = b, where b is the y-coordinate of the point (17/6, 5/2). Therefore, the equation of the perpendicular bisector is: y = 5/2 This is the equation of a line parallel to the x-axis and passing through the point (0, 5/2). Therefore, the locus of a point equidistant from the intersection of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a line parallel to 7x + 13y + 8 = 0, which is.