Start with a line segment connecting two points, A and B. [tex]\dfrac{DA}{BA}=3[/tex] 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 [tex]\dfrac{DB}{BA}=2[/tex], 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, [tex]\dfrac{AB}{CB}=\dfrac12[/tex], 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 [tex]d[/tex] be the length of AB. Then the length of CD in terms of [tex]d[/tex] is [tex]4d[/tex]. We have the coordinates of C and D, and the distance between them is [tex]\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2[/tex]. So
[tex]4d=4\sqrt2\implies d=\sqrt2[/tex]
The slope of the line through C and D is
[tex]\dfrac{0-4}{4-0}=-1[/tex]
and so the equation of the line through these points is
[tex]y-4=-(x-0)\implies x+y=4[/tex]
So the coordinates of A are [tex](x,y)=(x,4-x)[/tex]. The distance between C and A is [tex]d=\sqrt2[/tex], so we have
[tex]\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1[/tex]
Since A falls to the right of C (in the [tex]x,y[/tex] plane, not just in the sketch), we know to take the positive value [tex]x=1[/tex]. Then the [tex]y[/tex] coordinate is [tex]y=4-1=3[/tex].
All this to say that A is the point (1, 3), so
[tex]2x+y=2+3=5[/tex]
