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xz_007 [3.2K]
3 years ago
5

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

{cb} = \dfrac{1}{2}$ and the point $d$ is such that $\dfrac{da}{ba} = 3$ and $\dfrac{db}{ba} = 2.$ if $c = (0, 4),$ $d = (4, 0),$ and $a = (x, y),$ what is the value of $2x + y$?

Mathematics
1 answer:
Deffense [45]3 years ago
5 0

Start with a line segment connecting two points, A and B. \dfrac{DA}{BA}=3 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 \dfrac{DB}{BA}=2, 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, \dfrac{AB}{CB}=\dfrac12, 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 d be the length of AB. Then the length of CD in terms of d is 4d. We have the coordinates of C and D, and the distance between them is \sqrt{(4-0)^2+(0-4)^2}=4\sqrt2. So

4d=4\sqrt2\implies d=\sqrt2

The slope of the line through C and D is

\dfrac{0-4}{4-0}=-1

and so the equation of the line through these points is

y-4=-(x-0)\implies x+y=4

So the coordinates of A are (x,y)=(x,4-x). The distance between C and A is d=\sqrt2, so we have

\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1

Since A falls to the right of C (in the x,y plane, not just in the sketch), we know to take the positive value x=1. Then the y coordinate is y=4-1=3.

All this to say that A is the point (1, 3), so

2x+y=2+3=5

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