Geometry/Geometry

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Question
Find the locus of the points p(3x+2,2-x).

Answer
Let's change that to (3t+2, 2-t) and then call it (x,y).

If y = 2-t, then we need to solve for t(x).
Since we have x = 3t+2, 3t = x-2, so t = (x-2)/3.

Putting this in the equation for y gives y = 2 - (x-2)/3.
Multiplying out gives y = 2 - x/3 + 2/3.
Add 2 and 2/3 gives 8/3, so we have y = -x/3 + 8/3.

What we have is a line, slope -1/3 and y intercept of 8/3.
That is a line we know how to draw.

Two points would be at x=2 and x=5, since both of those give integers.
They would be (2,2) and (5,1).

From those two points, a line can be drawn.
Using the original problem, it be seen they are gotten by t=0 and t=1.  

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