# Geometry help on rotations.

1. Aug 7, 2014

### carojay

1. The problem statement, all variables and given/known data

1. A group of American physicists works on a project where planar lines are in the form
X=t⋅P+s⋅Q
where P , 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=t⋅P+s⋅Q in the following 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.

2. Relevant equations
I know for number 1, you basically just add the vector B.
for 2 and 3 I do not know whether to use point slope form and just change the slope or if I need to change the coordinates to (-y,x) for 90 degree rotation and (-x,-y) for 180 degree rotation but those are for rotation about the origin and my problem does not state that. Does the slope for a 180 degree rotation go back to the same slope? I am really confused on which direction to take.

3. The attempt at a solution
attempt at part 1:

2. Relevant equations

3. The attempt at a solution
Let a$,$b$, and$c$be fixed reals satisfying$a^2+b^2\ne 0$. They need to know formulae for the images of the line $$a\cdot x+b\cdot y+c=0$$\\ 1. Under the translation by a vector $$B=[u,v]$$,\\ First, solve for$y\to$$$y=\frac{-a}{b}\cdotx -\frac{c}{b}$$.\\ So, we know that the line crosses the y-axis at $$-\frac{c}{b}$$$\to$$$(0, -\frac{c}{b}$$.\\ The translation by a vector means that we add$B=[u,v]$,to yield\\ $$(0, -\frac{c}{b}) + (u,v)=(u,v-\frac{c}{b})$$$\to$$$(u,\frac{bv-c}{b})$$\\ Now we let$x_0=u, y_0=(\frac{bv-c}{b})\$\\
$$y-y_0=m(x-x_0) \to y-(\frac{bv-c}{b})=mx – mu$$\\
$$-m\cotx + y = \frac{bv-c}{b}) –mu \to -m\cotx + y - \frac{bv-c}{b})+mu=0$$\\

2. Aug 7, 2014

### Simon Bridge

X=tP+sQ is actually a plane, hence the restriction that t+s=1 ... so you can rewrite:

X=tP+(1-t)Q = Q+(P-Q)t ... which is a more familiar form of the line.

Basically P and Q are both on the line, and P-Q is a vector pointing along the line.
You know how to rotate a vector?

I'd be thinking in terms of the vector Q-C ... may help to draw a sketch.

3. Aug 8, 2014

### Fredrik

Staff Emeritus