In a coordinate plane, the points having an x-coordinate of 1 are the points (1,y) for arbitrary "y". So, the points you are looking for, satisfy this distance equation= 5, or = 5. Square both sides to get 16 + (y+2)^2 = 5^2 (y+2)^2 = 25 - 16 (y+2)^2 = 9 y + 2 = = +/- 3 Hence, there are two values for y = 3 - 2 = 1 and = -3 - 2 = -5. So, the points are A = (1,1) and B = (1,-5). ANSWER