Question 399366
 A (4 , 1), B (6 , 1) , C (4 , 6).
In a triangle A'B'C', A' (0 , -1), B' (-2 , -1), C' (0 , -6). 
<pre><font face = "batangche" color = "indigo" size = 4><b>
{{{drawing(400,400,-7,7,-7,7,green(triangle(0,-1,-2,-1,0,-6)),
blue(line(0,-1,0,-6)),
locate(4,1,A), locate(6,1,B),locate(4,6.5,C), locate(0.4,-.7,"A'"),
locate(-2.7,-1,"B'"),locate(.4,-5.5,"C'"),

red(triangle(4,1,6,1,4,6)), graph(400,400,-7,7,-7,7)  )}}}

Let's connect corresponding points:

{{{drawing(400,400,-7,7,-7,7,green(triangle(0,-1,-2,-1,0,-6)),
blue(line(4,1,0,-1),line(6,1,-2,-1), line(4,6,0,-6)),
locate(4,1,A), locate(6,1,B),locate(4,6.5,C), locate(0.4,-.7,"A'"),
locate(-2.7,-1,"B'"),locate(.4,-5.5,"C'"), locate(1.7,.5,P),



red(triangle(4,1,6,1,4,6)), graph(400,400,-7,7,-7,7)  )}}}




To find the center or rotation find the midpoint P of either

AA', BB', or CC'

If you find the midpoint of AA' using the formula

M = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

M = ({{{(4+0)/2}}}, {{{(1+(-1))/2}}})

M = ({{{4/2}}}, {{{0/2}}})
 
M = (2,0), that's the center of rotation.

It is not necessariy to do but one, but as a check,

If you find the midpoint of BB' using the formula

M = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

M = ({{{(6+(-2))/2}}}, {{{(1+(-1))/2}}})

M = ({{{4/2}}}, {{{0/2}}})
 
M = (2,0), that gives the same center of rotation.

and

If you find the midpoint of CC' using the formula

M = ({{{(x[1]+x[2])/2}}}, {{{(y[1]+y[2])/2}}})

M = ({{{(4+0)/2}}}, {{{(6+(-6))/2}}})

M = ({{{4/2}}}, {{{0/2}}})
 
M = (2,0), which also gives the same center of rotation.

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The angle of rotation is 180°, because angles APA', BPB' and CPC'
are all straight angles, or 180°.

Edwin</pre>