Nemesis is back! Appreciate any simpler methods! Layman terms if possible.

Is the Answer (a) y = x --> (1,1), (2,2) So contains (b) y = x + 0.5 -> y - x = 0.5 Not possible (c) y = x + 5 -> y - x = 5 --> (1,6) (2,7) So contains (d) y = 0.5x -> (2,1) (4,2) So contains (e) y = 0.5x + 5 -> (2,6) (4,7) So contains

Which of the following lines in the xy-plane does not contain any point with integers as both coordinates? A) y = x B) y = x + 1/2 C) y = x + 5 D) y = (1/2)x E) y = (1/2)x + 5 If a point (x,y) is on a given line, then the coordinates (x and y) of that point must satisfy the equation of that line. In answer choice B, the equation (y = x + 1/2) tells us that, for ANY POINT ON THE LINE, the y-coordinate will be equal to the x-coordinate plus 1/2 So, if the x-coordinate is an integer, the y-coordinate cannot be an integer. So, it's impossible for any point on this line to have integers as both coordinates. Answer: B Cheers, Brent

vinay1983 wrote: Which of the following lines in the xy-plane does not contain any point with integers as both coordinates? (A) y = x (B) y = x + 1/2 (C) y = x + 5 (D) y = x*1/2 (E) y = x/2 + 5 We see that the point (1,1) is on the line y = x (answer choice A); (1, 6) on y = x + 5 (C); (2, 1) on y = x*1/2 (D); and (2, 6) on y = x/2 + 5 (E). Thus, the only line that won’t have any point with integers as both coordinates is y = x + 1/2 (answer choice B). The reason is simple: if x is an integer, then y can’t be an integer, since the sum of an integer and 1/2 will never be an integer. Similarly, if y is an integer, then x can’t be an integer, since the difference of an integer and 1/2 (notice that x = y - 1/2) can never be an integer. Answer: B