Find four points that are 15 units away from the point (4,9). consider the fact that 3, 4, 5 is a pythagorean triple. select all that apply. (16, 18) (-8, 0) (16, 21) (13, 21) (-5, -3) (0, 13)

To find points that are 15 units away, first we need to know the solution to the hypotenuse in question. (x, y) = (4, 9) where x is the adjacent and y is opposite. From the Pythagorean Triple. X^2 = 4^2 + 9^2 = 97. Hence X = âš97 = 9.848 unit. To find points 15 units away, we seek hypotenuse that are 24 units in length. Note: 15units will be added to the original hypotenuse, which is 9.848. The first hypotenuse is X = âš16^2 + 18^2 = 24.08 which is one of the points. We need three more. The second X = âš0^2 + 8^2 = 8. So the second is out of the race. The third X = âš16^2 + 21^2 = 26.4 unit; this is 17 units away and not 15units, so this is out too. The fourth X =âš13^2 + 21^2 = 24.68 units. Yes, this makes it two. The fifth X = âš(-5)^2 + (-3)^2 = 5.83; this is out too. And finally X = âš0^2 + 13^2 = 13. Strictly speaki2, there are only two points that are 24 units away (16, 18) and (13, 21). The third point (16, 21) exceeds 24 unit.

Ithink the answer is b