# Find the formula for X=tP+sQ under a translation and a rotation

1. Aug 7, 2014

### KitKat21

1. The problem statement, all variables and given/known data
A group of American physicist works on a project where planar lines are in the form X=tP+sQ, where P and Q are two fixed different points and s, t are varying reals satisfying s+t=1. They need to know formulae for the images of the line X=tP+sQ in the following three cases:
1. Under the translation by a vector B
2. Under rotation about a point C by 180 degrees
3. Under rotation about a point C by 90 degrees
Please provide those formulae and a justification for them.

2. Relevant equations
X=tP+sQ
s+t=1

3. The attempt at a solution
A translation by a vector, B will preserve length and slope, so the new formulae is X=tP+sQ+B.

A rotation will preserve length, but not necessarily slope. I know that with points, a 90 degree rotation will give (x,y) -> (y,-x) and that a rotation of 180 degrees will give (x,y) -> (-x,y).

I'm not sure how to use this information to get to a clear answer, with a formula and justification for it.

2. Aug 7, 2014

### Ray Vickson

There are standard formulas for the transition from $(x,y)$ to $(x',y')$ under a rotation through angle $\theta$. Just use them on each of the points P and Q.

3. Aug 7, 2014

### KitKat21

Where would I find these formulas? I have tried google and come up with nothing...?

4. Aug 8, 2014

### verty

For a rotation at C, translate C to the origin, rotate, translate the origin to C.