Answer to Question #236241 in Analytic Geometry for luka

Question

Given point A (2 , -3) in a plane; find the coordinates of its image:

a) Under enlargement about ( -3 , 1 )with scale factor 2; (3marks)

b) Under rotation of center ( -2 , 1 ) and angle θ=π/6

Answer 1

a) Distance from centre of enlargement $( -3 , 1 )$ to point $A (2 , -3)$ is is multiplied by scale factor 2

$(x_1-(-3))=2(2-(-3))=>x_1=7$

$(y_1-1)=2(-3-1)=>y_1=-7$

$A_1(7, -7)$

b)

$(x+2)^2+(y-1)^2=32$

$y=-x-1$

Points $A(2,-3)$ and $A_2(x_2, y_2)$ lie on the circle

$(x+2)^2+(y-1)^2=32$

Point $A(2,-3)$ is the point of intersection this circle and the line

$y=-x-1$

Rotation of center $( -2 , 1 )$ and angle $θ=π/6.$

The image of the line $y=-x-1$ is the line

$y=-\tan(\dfrac{\pi}{4} -\dfrac{\pi}{6})x+b$

$y=-\tan(\dfrac{\pi}{12})x+(1-2\tan(\dfrac{\pi}{12}))$

Point $A_2(x_2,y_2)$ is the point of intersection the circle

$(x+2)^2+(y-1)^2=32$

and the line

$y=-\tan(\dfrac{\pi}{12})x+(1-2\tan(\dfrac{\pi}{12}))$

Then

$y_2=-\tan(\dfrac{\pi}{12})x_2+(1-2\tan(\dfrac{\pi}{12}))$

$(x_2+2)^2+(y_2-1)^2=32$

Substitute

$(x_2+2)^2+(-\tan(\dfrac{\pi}{12})x_2-2\tan(\dfrac{\pi}{12}))^2=32,$

$x_2>0$

$x_2^2+4x_2+4+\tan^2(\dfrac{\pi}{12})x_2^2+4\tan(\dfrac{\pi}{12})x_2$

$+4\tan^2(\dfrac{\pi}{12})=32$

$\sec^2(\dfrac{\pi}{12})x_2^2+4(1+\tan(\dfrac{\pi}{12}))x_2$

$+4\sec^2(\dfrac{\pi}{12})-32=0$

$D=16+32\tan(\dfrac{\pi}{12})+16\tan^2(\dfrac{\pi}{12})-16\sec^4(\dfrac{\pi}{12})$

$+128\sec^2(\dfrac{\pi}{12})$

Since $x_2>0$

$x_2=-2\cos^2(\dfrac{\pi}{12})+\cos^2(\dfrac{\pi}{12})\sqrt{\dfrac{D}{4}}$

$y_2=-\tan(\dfrac{\pi}{12})x_2+(1-2\tan(\dfrac{\pi}{12}))$