# Geometry/Geometry

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

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.

Geometry

Volunteer

#### Scott A Wilson

##### Expertise

I can answer whatever questions you ask except how to trisect an angle. The ones I can answer include constructing parallel lines, dividing a line into n sections, bisecting an angle, splitting an angle in half, and almost anything else that is done in geometry.

##### Experience

I have been assisting people in Geometry since the 80's.

Education/Credentials
I have an MS at Oregon State and a BS at Oregon State, both with honors.

Awards and Honors
I was the outstanding student in high school in the area of geometry and math in general.

Past/Present Clients
Over 8,500 people, mostly in math, with almost 450 in geometry.