The two points on the line are (1,-2) and (4,5). From these two lines, we can deduce the equation of the line.
So, (y+2)/(5+2) = (x-1)/(4-1) which gives the equation as 7x - 3y = 13.
Let the point be (h,k). THis point and the point (4,5) are equidistant from the centre (1,-2).
Let the other point be (h,k). So putting the distance between the two point equation, we have
(4-1)2 + (5+2)2 = (h-1)2 + (k+2)2
Substituting k = (7h-13)/3 in the above equation , and solving for h , we have two values of j
h = 3 and -3.
Solving for k we have two values of k. k = 5, -9.
So the two points are (4,5) and (-2,-9).
Since (4,5) is already one end point, other other point is (-2,-9)