Question

Please dont use Trig to solve problem Find the locus of points \( X \) such that \( A X=3 B X \), where \( A=(0,0) \) and \( B=(1,0) \).

Answer 1

• To find the locus of points where the distance from point A is three times the distance from point B, we first need to analyze the positioning of the points.
  
  Initial Points:
  1. A(0,0)
  2. B(1,0)
  
  Now, let's start solving this step by step:
  
  1. In order for a point to be equidistant from two separate points (in this case A and B), it must lie on the perpendicular bisector of the line segment joining the two points.
  
  2. Finding the Midpoint: The midpoint (M) of AB is the average of the x-coordinates and the y-coordinates of A and B respectively. So, M is ((0 + 1)/2, (0 + 0)/2) = (0.5, 0).
  
  3. Understanding the Perpendicular Bisector: The perpendicular bisector of a line segment is the line that passes through the midpoint of the line segment and is perpendicular to it. The slope of AB (line segment joining A and B) can be defined as the change in 'y' divided by the change in 'x'. But as both points are on the x-axis so the slope of AB is 0.
  
  4. Since the slope of AB is 0, the slope of the line perpendicular to AB is undefined, or in other words, the line is vertical.
  
  5. Now, the vertical line crossing through the point (0.5, 0) is the perpendicular bisector of the line joining A and B.
  
  Therefore, the locus of points X such that AX = 3BX is the line x = 0.5. Any point on this line will satisfy the given conditions. Therefore, the locus of points is described by the set "All points (0.5, y) for any y."