Answer to this question is B, not C.
A rectangle is plotted on the standard coordinate plane, with vertices at the origin and (0,6). If the coordinates of all vertices of the rectangle are non-negative integers, what are the coordinates of the other two vertices?Notice that we are told that the coordinates of
all vertices of the rectangle are
non-negative integers. That basically means that the rectangle is plotted entirely in the first quadrant: one vertex at the origin (0, 0), another on Y-axis (0, 6), third one somewhere in the first quadrant (x, 6) and the fourth vertex on the X-axis (x, 0):
Attachment:
Rectangle.png [ 5.05 KiB | Viewed 18476 times ]
(1) The distance between the origin and one of the other vertices is 10 units. 10 units is either the length of the other side or the length of the diagonal. The vertices could be: (0, 0), (0, 6), (10, 6) and (10, 0) OR (0, 0), (0, 6), (8, 6) and (8, 0), in this case the distance of 10 units is the distance between (0, 0) and (6, 8). Not sufficient.
(2) The distance between the origin and one of the other vertices is 8 units. Now, 8 units can not be the length of the diagonal since in this case the coordinates of the other two vertices won't be
integers: if the diagonal=8, then the length of the other side of the rectangle is \(\sqrt{8^2-6^2}=2\sqrt{7}\), so the coordinates of the other two vertices are: \((2\sqrt{7}, \ 6)\) and \((2\sqrt{7}, \ 0)\). So, 8 units must be the length of the side, therefore the vertices are (0, 0), (0, 6), (8, 6) and (8, 0). Sufficient.
Answer: B.
Hope it's clear.
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