Find a point m on the straight line L: 3x-y-1 = 0 so that the distance difference between it and a (4,1) and B (0,4) is the largest, and find the maximum value

Find a point m on the straight line L: 3x-y-1 = 0 so that the distance difference between it and a (4,1) and B (0,4) is the largest, and find the maximum value

(1) First, find out the symmetric point B '(3,3) of point B (0,4) with respect to the line 3x-y-1 = 0. (2). Connect point ab' and extend it. The intersection line L: 3x-y-1 = 0 is at point m (2,5). It can be proved that point m is the required point, and the maximum distance difference is | ab '| = √ 5

On the straight line L: 3x-y-1 = 0, find a point P so that the distance difference between it and a (4,1), B (0,4) is the largest

From the known ABP three points are collinear (the difference between the two sides of the triangle is less than the third side)
The solution can be obtained by solving the equations of AB line and L line

Find a point on the line x + 2Y = 0 so that its distance to the origin is equal to that of the line x + 2Y - 5 = 0

Let this point be (- 2Y, y)
Use the formula of distance from point to line d = absolute value (- 2Y + 2y-5) / √ 5 = √ 5
So it's √ ((- 2Y) ^ 2 + y ^ 2) = √ 5
y=±1
So (- 2,1) and (2, - 1)