3(4 + Зр) How do u solve this?
1 answer:
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Step-by-step explanation:
here's your answer hope it helps you
Answer:
A(2, 2)
Step-by-step explanation:
I find it useful to graph the given points. The center of rotation is at the place where the perpendicular bisectors of BB' and CC' meet. That point is ...
A = (2, 2).
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The graph shows you the slope of BB' is 1, so its perpendicular bisector will have a slope of -1. The midpoint of BB' is (2-2, 6+2)/2 = (0, 4). This is the y-intercept of the line, so the perpendicular bisector of BB' has equation ...
y = -x +4
The slope of CC' is -1/3, so its perpendicular will have a slope of -1/(-1/3) = 3. The midpoint of CC' is (4+1, 3+4)/2 = (5/2, 7/2). In point-slope form the equation of the perpendicular bisector of CC' is ...
y -7/2 = 3(x -5/2)
2y -7 = 3(2x -5) . . . . multiply by 2
2(-x+4) -7 = 3(2x -5) . . . substitute for y
-2x +1 = 6x -15 . . . . .eliminate parentheses
16 = 8x . . . . . . . . . . add 2x+15
x = 2 . . . . . . divide by 8
y = -2+4 = 2 . . . . from the equation for y
The intersection of the perpendicular bisectors of BB' and CC' is the center of rotation:
A = (2, 2)
