Since we know the midpoint is equidistant from the two endpoints, we know that the x value of one of the endpoints will be a certain distance to the right of (2,6) and the other endpoint will be the same distance to the left of it. The same applies for the y direction, one will be a certain distance above the midpoint, the other will be the same distance below it. After graphing the equations we can see that the equation y = 2x will be the equation to the right and above. Therefore if we say that the distance to the right of the midpoint equals x, then the distance to the left equals -x, if the distance above we call y, then the distance below is -y. Now we can rewrite the endpoints as follows:
(2+x, 6+y) and (2-x, 6-y)
We then substitute these values into our two equations:
(6 + y) = 2(2 + x) and (6 - y) = (2 - x) + 3
6 + y = 2x + 4 and 6 - y = 5 - x
y = 2x - 2 and y = x + 1
Setting the 2 equations equal to each other, we get:
2x - 2 = x + 1
x = 3
Since y = x - 1
y = 4
Therefore one point is 3 right and 4 up from the midpoint or (2+3, 6+4) = (5, 10)
and the other point is 3 left and 4 down from the midpoint or (2-3, 6-4) = (-1, 2)
Joe P.
tutor
Vivian,
You are correct that the two graphs intersect at (3,6) but that doesn't prevent the possibility of a midpoint between to points on those lines being somewhere else.
For example, graph y = 0 and x = 0 our two axis lines. They intersect at (0,0). However, the point (1,1) could be a midpoint between the two lines, for points (2, 0) and (0, 2). Point (2, 2) could also be a midpoint between (4, 0) and (0, 4). There is actually an infinite number of possible midpoints for my example and the all lie on the y = x line. Your problem also has an infinite number of midpoints between the two lines, (2, 6) is just one of them.
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02/03/14
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02/02/14