a) Distance from centre of enlargement ( − 3 , 1 ) ( -3 , 1 ) ( − 3 , 1 ) to point A ( 2 , − 3 ) A (2 , -3) A ( 2 , − 3 ) is is multiplied by scale factor 2
( x 1 − ( − 3 ) ) = 2 ( 2 − ( − 3 ) ) = > x 1 = 7 (x_1-(-3))=2(2-(-3))=>x_1=7 ( x 1 − ( − 3 )) = 2 ( 2 − ( − 3 )) => x 1 = 7
( y 1 − 1 ) = 2 ( − 3 − 1 ) = > y 1 = − 7 (y_1-1)=2(-3-1)=>y_1=-7 ( y 1 − 1 ) = 2 ( − 3 − 1 ) => y 1 = − 7 A 1 ( 7 , − 7 ) A_1(7, -7) A 1 ( 7 , − 7 )
b)
( x + 2 ) 2 + ( y − 1 ) 2 = 32 (x+2)^2+(y-1)^2=32 ( x + 2 ) 2 + ( y − 1 ) 2 = 32
y = − x − 1 y=-x-1 y = − x − 1
Points A ( 2 , − 3 ) A(2,-3) A ( 2 , − 3 ) and A 2 ( x 2 , y 2 ) A_2(x_2, y_2) A 2 ( x 2 , y 2 ) lie on the circle
( x + 2 ) 2 + ( y − 1 ) 2 = 32 (x+2)^2+(y-1)^2=32 ( x + 2 ) 2 + ( y − 1 ) 2 = 32 Point A ( 2 , − 3 ) A(2,-3) A ( 2 , − 3 ) is the point of intersection this circle and the line
y = − x − 1 y=-x-1 y = − x − 1 Rotation of center ( − 2 , 1 ) ( -2 , 1 ) ( − 2 , 1 ) and angle θ = π / 6. θ=π/6. θ = π /6.
The image of the line y = − x − 1 y=-x-1 y = − x − 1 is the line
y = − tan ( π 4 − π 6 ) x + b y=-\tan(\dfrac{\pi}{4} -\dfrac{\pi}{6})x+b y = − tan ( 4 π − 6 π ) x + b
y = − tan ( π 12 ) x + ( 1 − 2 tan ( π 12 ) ) y=-\tan(\dfrac{\pi}{12})x+(1-2\tan(\dfrac{\pi}{12})) y = − tan ( 12 π ) x + ( 1 − 2 tan ( 12 π ))
Point A 2 ( x 2 , y 2 ) A_2(x_2,y_2) A 2 ( x 2 , y 2 ) is the point of intersection the circle
( x + 2 ) 2 + ( y − 1 ) 2 = 32 (x+2)^2+(y-1)^2=32 ( x + 2 ) 2 + ( y − 1 ) 2 = 32 and the line
y = − tan ( π 12 ) x + ( 1 − 2 tan ( π 12 ) ) y=-\tan(\dfrac{\pi}{12})x+(1-2\tan(\dfrac{\pi}{12})) y = − tan ( 12 π ) x + ( 1 − 2 tan ( 12 π ))
Then
y 2 = − tan ( π 12 ) x 2 + ( 1 − 2 tan ( π 12 ) ) y_2=-\tan(\dfrac{\pi}{12})x_2+(1-2\tan(\dfrac{\pi}{12})) y 2 = − tan ( 12 π ) x 2 + ( 1 − 2 tan ( 12 π ))
( x 2 + 2 ) 2 + ( y 2 − 1 ) 2 = 32 (x_2+2)^2+(y_2-1)^2=32 ( x 2 + 2 ) 2 + ( y 2 − 1 ) 2 = 32 Substitute
( x 2 + 2 ) 2 + ( − tan ( π 12 ) x 2 − 2 tan ( π 12 ) ) 2 = 32 , (x_2+2)^2+(-\tan(\dfrac{\pi}{12})x_2-2\tan(\dfrac{\pi}{12}))^2=32, ( x 2 + 2 ) 2 + ( − tan ( 12 π ) x 2 − 2 tan ( 12 π ) ) 2 = 32 ,
x 2 > 0 x_2>0 x 2 > 0
x 2 2 + 4 x 2 + 4 + tan 2 ( π 12 ) x 2 2 + 4 tan ( π 12 ) x 2 x_2^2+4x_2+4+\tan^2(\dfrac{\pi}{12})x_2^2+4\tan(\dfrac{\pi}{12})x_2 x 2 2 + 4 x 2 + 4 + tan 2 ( 12 π ) x 2 2 + 4 tan ( 12 π ) x 2
+ 4 tan 2 ( π 12 ) = 32 +4\tan^2(\dfrac{\pi}{12})=32 + 4 tan 2 ( 12 π ) = 32
sec 2 ( π 12 ) x 2 2 + 4 ( 1 + tan ( π 12 ) ) x 2 \sec^2(\dfrac{\pi}{12})x_2^2+4(1+\tan(\dfrac{\pi}{12}))x_2 sec 2 ( 12 π ) x 2 2 + 4 ( 1 + tan ( 12 π )) x 2
+ 4 sec 2 ( π 12 ) − 32 = 0 +4\sec^2(\dfrac{\pi}{12})-32=0 + 4 sec 2 ( 12 π ) − 32 = 0 D = 16 + 32 tan ( π 12 ) + 16 tan 2 ( π 12 ) − 16 sec 4 ( π 12 ) D=16+32\tan(\dfrac{\pi}{12})+16\tan^2(\dfrac{\pi}{12})-16\sec^4(\dfrac{\pi}{12}) D = 16 + 32 tan ( 12 π ) + 16 tan 2 ( 12 π ) − 16 sec 4 ( 12 π )
+ 128 sec 2 ( π 12 ) +128\sec^2(\dfrac{\pi}{12}) + 128 sec 2 ( 12 π ) Since x 2 > 0 x_2>0 x 2 > 0
x 2 = − 2 cos 2 ( π 12 ) + cos 2 ( π 12 ) D 4 x_2=-2\cos^2(\dfrac{\pi}{12})+\cos^2(\dfrac{\pi}{12})\sqrt{\dfrac{D}{4}} x 2 = − 2 cos 2 ( 12 π ) + cos 2 ( 12 π ) 4 D
y 2 = − tan ( π 12 ) x 2 + ( 1 − 2 tan ( π 12 ) ) y_2=-\tan(\dfrac{\pi}{12})x_2+(1-2\tan(\dfrac{\pi}{12})) y 2 = − tan ( 12 π ) x 2 + ( 1 − 2 tan ( 12 π ))
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