Assume P(xp,yp), A(xa,ya), etc. We know that rotation rule of 90<span>° clockwise about the origin is R_-90(x,y) -> (y,-x) For example, rotating A about the origin 90</span><span>° clockwise is (xa,ya) -> (ya, -xa) or for a point at H(5,2), after rotation, H'(2,-5), etc.
To rotate about P, we need to translate the point to the origin, rotate, then translate back. The rule for translation is T_(dx,dy) (x,y) -> (x+dx, y+dy)
So with the translation set at the coordinates of P, and combining the rotation with the translations, the complete rule is: T_(xp,yp) R_(-90) T_(-xp,-yp) (x,y) -> </span>T_(xp,yp) R_(-90) (x-xp, y-yp) -> T_(xp,yp) (y-yp, -(x-xp)) -> (y-yp+xp, -x+xp+yp)
Example: rotate point A(7,3) about point P(4,2) => x=7, y=3, xp=4, yp=2 => A'(3-2+4, -7+4+2) => A'(5,-1)
Let the speed for the first 12 mi be x mi/h, the speed for 18 mi was (x+4) mi/h thus given that time=distance/speed the average time will be: 3=(12+18)/(x+x+4) 3=30/(2x+4) solving for x we get 3(2x+4)=30 6x+6=30 6x=24 x=4 mi/hr Answer: 12 mi/hr
