Start with a line segment connecting two points, A and B.
means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.
But with
, or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.
Finally,
, so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.
Now let
be the length of AB. Then the length of CD in terms of
is
. We have the coordinates of C and D, and the distance between them is
. So
![4d=4\sqrt2\implies d=\sqrt2](https://tex.z-dn.net/?f=4d%3D4%5Csqrt2%5Cimplies%20d%3D%5Csqrt2)
The slope of the line through C and D is
![\dfrac{0-4}{4-0}=-1](https://tex.z-dn.net/?f=%5Cdfrac%7B0-4%7D%7B4-0%7D%3D-1)
and so the equation of the line through these points is
![y-4=-(x-0)\implies x+y=4](https://tex.z-dn.net/?f=y-4%3D-%28x-0%29%5Cimplies%20x%2By%3D4)
So the coordinates of A are
. The distance between C and A is
, so we have
![\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1](https://tex.z-dn.net/?f=%5Csqrt%7B%28x-0%29%5E2%2B%284-x-4%29%5E2%7D%3D%5Csqrt%7B2x%5E2%7D%3D%7Cx%7C%5Csqrt2%3D%5Csqrt2%5Cimplies%7Cx%7C%3D1)
Since A falls to the right of C (in the
plane, not just in the sketch), we know to take the positive value
. Then the
coordinate is
.
All this to say that A is the point (1, 3), so
![2x+y=2+3=5](https://tex.z-dn.net/?f=2x%2By%3D2%2B3%3D5)