Answers
1) The coordinates of A' are (4,1).
2) The coordinates of A″ are (11,2).
Solution
1) The ∆ABC rotates an angle A=180° counterclocwise around point D to create ∆A′B′C′
The coordinates of A are (4,9)=(xa,ya)→xa=4, ya=9
The coordinates of A with respect the point of rotation D(4,5)=(xd,yd)→xd=4, yd=5 are (xa-xd, ya-yd)=(4-4,9-5)=(0,4)=(x,y)→x=0, y=4
The coordinates of A'=(x',y') with respect the point of rotation D are:
x'= x cos A - y sin A
x' = 0 cos 180° - 4 sin 180°
x'= 0 (-1) - 4 (0)
x'=0-0
x'=0
y' = y cos A + x sin A
y' = 4 cos 180° + 0 sin 180°
y' = 4 (-1) + 0 (0)
y' = -4+0
y'= -4
(x',y')=(0,-4)
The coordinates of A'=(xa', ya') with respect the origin are:
xa'=x'+xd→xa'=0+4→xa'=4
ya'=y'+yd→ya'=-4+5→ya'=1
The coordinates of A' are (xa',ya')=(4,1)
2) The ∆A′B′C′ rotates an angle B=90° counterclockwise around point E(7, 5) to form triangle ∆A″B″C″
The coordinates of A' are (4,1)=(xa',ya')→xa'=4, ya'=1
The coordinates of A' with respect the point of rotation E(7,5)=(xe,ye)→xe=7, ye=5 are (xa'-xe, ya'-ye)=(4-7,1-5)=(-3,-4)=(x,y)→x=-3, y=-4
The coordinates of A''=(x'',y'') with respect the point of rotation E are:
x''= x cos B - y sin B
x'' = -3 cos 90° - (-4) sin 90°
x''= -3(0) + 4 (1)
x''= -0+4
x''=4
y'' = y cos B + x sin B
y'' = -4 cos 90° + (-3) sin 90°
y'' = -4 (0) - 3 (1)
y'' = -0-3
y''= -3
(x'',y'')=(4,-3)
The coordinates of A''=(xa'', ya'') with respect the origin are:
xa''=x''+xe→xa''=4+7→xa''=11
ya''=y''+ye→ya''=-3+5→ya''=2
The coordinates of A'' are (xa'',ya'')=(11,2)
