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Consider point P(x, y) in first quadrant. Its reflection about x-axis is Q(x_(1), y_(1)). So, x_(1)=x and y(1)=-y. This may be written as : {(x_(1)=1. x+0.y),(y_(1)=0. x+(-1)y):} This system of equations can be put in the matrix as : [(x_(1)),(y_(1))]=[(1,0),(0,-1)][(x),(y)] Here, matrix [(1,0),(0,-1)] is the matrix of reflection about x-axis. Then find the matrix of (i) reflection about y-axis (ii) reflection about the line y=x (iii) reflection about origin (iv) reflection about line y=(tan theta)x |
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Answer» Solution :(i) Reflection of (X, y) about y-axis is `(x_(1), y_(1)) equiv (-x, y)`. `:. x_(1)=(-1)x+0y` and `y_(1)=0x+y` `:. [(x_(1)),(y_(1))]=[(-1,0),(0,1)][(x),(y)]` (ii) Reflection of (x, y) about the line `y=x` is `(x_(1), y_(1)) equiv (y, x)`. `:. x_(1)=0x+y` `y_(1)=x+0y` `:. [(x_(1)),(y_(1))]=[(0,1),(1,0)][(x),(y)]` (III) Reflection of (x, y) about origin is `(x_(1), y_(1)) equiv (-x, -y)`. `:. x_(1)=-x+0y` `y_(1)=0x-y` `:. [(x_(1)),(y_(1))]=[(-1,0),(0,-1)][(x),(y)]` (iv) Reflecton in line `t=x tan theta` or `(SIN theta)x-(cos theta)y=0`:  We kanow that `(x_(1)-x)/(sin theta)=(y_(1)-y)/(- cos theta)=(-2((sin theta)x-(cos theta)y))/(sin^(2) theta+cos^(2) theta)` `:. x_(1)=(1-2 sin^(2) theta)x+(2 sin theta cos theta)y` or `x_(1)=(cos 2 theta)x+(sin 2 theta) y` `y_(1)=(2 sin theta cos theta)x+ (1-2 cos^(2) theta)y` or `y_(1)=(sin 2 theta)x-(cos 2 theta)y` Thus, `[(x_(1)),(y_(1))]=[(cos 2 theta,sin 2 theta),(sin 2 theta,- cos 2 theta)][(x),(y)]` By putting `theta=0, pi//2, pi//4`, we can GET the reflection matrices x-axis, y-axis and the line y=x, respectively. |
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