Thus line of reflection pass through the origin the transformation is linear and could be expressed by 2 x 2 matrix. [br]1. Reflect base vectors e1, e2.[br]2. Images e1' and e2' are columns of the matrix of linear transformation.[br]3. Check the representation by means of the picture "lion".

Method 1 (Experimental): Move with point [color=#0000ff][i]B[/i][/color] to get location, where[color=#0000ff][i] B = B'[/i][/color]. Discover all such position. [br]Method 2 (Algebraic): Matrix equation of transformation[color=#0000ff][i] X' = MX t[/i][/color]ogether with condition for fixed points [i][color=#0000ff]X' = X[/color][/i] give equation [color=#0000ff][i]X = MX[/i][/color] for unknown fixed point [color=#0000ff][i]X[/i][/color]. Rearrangement in homogenous form:[br][math]\left(M-E\right)X=o[/math][br]could be solved by Gauss elimination. GeoGebra command [code]ReducedRowEchelonForm(matrix)[/code][br] of the matrix [color=#0000ff]([i]M - E[/i])[/color] returns the equivalent echelon form. [br]Matrix for reflection in line [i][color=#0000ff]OA[/color][/i], [color=#0000ff]O=(0,0), A =(1, 2)[/color] is [code]M = {{-0.6, 0.8},{0.8, 0.6}}[/code], thus[center][br][img]data:image/png;base64,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[/img][br][/center]Using back-substitution, unknown coordinates [color=#0000ff][i]x, y[/i][/color] can be solved for.[br][color=#0000ff][i]x - 0.5 y = 0[/i][/color], hence [color=#0000ff][i]y = 2x[/i][/color].

Method 1 (experimental): Move with dynamic points [color=#0000ff][i]B, C[/i][/color] and investigate the relationship of line [color=#0000ff][i]BC[/i][/color] and its image [color=#0000ff][i]B'C'[/i][/color]. We look for position where line [color=#0000ff][i]BC[/i][/color] is parallel with [i][color=#0000ff]B'C'[/color][/i].[br][br]Method 2 (algebraic): Fixed direction (=eigenvectors) [color=#0000ff][i]X[/i][/color] fulfils equation [i][color=#0000ff]X' = λX[/color][/i] for some [color=#0000ff]λ[/color], i.e. we should find coefficient [color=#0000ff]λ[/color] for which [i][color=#0000ff]λX = MX[/color][/i] has nonzero solution [color=#0000ff][i]X[/i][/color]. Rearrangement in homogenous form [color=#0000ff](M-λE)X = o[/color] gives the necessary condition for nonzero solution: [center][color=#0000ff]det(M-λE) = 0[/color][br][img]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAMoAAABzCAYAAADZoJ0VAAAGXElEQVR4nO2d3ZWjOBBGSWDzmBjIYpNQHpOBotHTZqJgah+g3RhjuySVkAT3ntMP0+1uajAX/fpjEgD4ytS6AIARQBQABYgCoABRABQgCoACRAFQgCgAChAFQAGiAChAFAAFiAKgAFEAFHQtSgxefIitywADRn8vuxUlOCcutK4CLInBifNjytKlKNEjyWUJTuYBZelPlOhlLrQk+lmmaZJpmpKE+/09J6aexiBunmSaZml5jaSel+Cmx+snwztXcG3PQw6diRLFu8KLNLjNhR7EqS6K5XXmgojI8n/yEtfamnU9Es9LcJNM81r3z7+tZIle5s3fHoG+RIm+sFmO4ufdBRDc0xv+yirJGX296MU36VKmnpfl9U/vxdfzmMZorUpXokRfePKil3nfvTn63oale1GjJTk6VqOLI/m8LKJsxYh+th1bDDZW6UiUKH627HY9vvmhm7H8bPZ+7XrVk2YZH5wj5AvJ50VWkX5a2iDOvKtU42/WoyNRgrjS7k/qBRGcTNMk8+zX31nvpOaD+eXuPTs/jiiP36s4diu9MZ5IP6JE/3Gg+zQDc/TlwvEFsd4Z34uye/16cWxfrzr2+8rXCzJK8I3uoKnnZXnBeiHXm+gYaZzSjygWfdbUvnjunVZf0GZQHMW3Gclnj92278d+FsymLIcoyQSLRcbX2Z3o5w9v8IEUXy6gFJ4vriC+2VVhcF4kiDNuVaKfh1lYvpgospvG/N467AfZZusF+9Yq+nUMFiW0uJMmnZfXWS/r6WERRMnDShT5tAJ9cAFIhRXow1YprBMFs7hGmwPTzsvPxMb6VWGGClFyMBQFxuB+onydQVHw94/8+WtSDQzCf//+Y/qe5+7x01AuymOuvbA4WpTbYdqiZO3x03Nyi3Kw5+gHRLkddqLk7PFL42RRPq+SI8q9MBMlY50olWxRjlarf//TzzMmy8JV2OynOljpRZTbYSZK9YXjTFEOP6vwKCqKnzcmb1ubL9tJEOVeHImStV2oS1HeNnNrUZvB/fZr9hFR4ImqLYrFTOyGZFEOt4vvRXk3iEIU2HDtMUpwr5/9fmlR3uwJQhTYYNb1St7Llk7GGOVg2/WTAMf7hOh6wR7zdZSEPX6pZM567Wew9jNfu58/Kt7OhjHrdXest7D0vTJvBaLcjvvt9bLgpL1evWRbWTB6fpn1Xq+a3EuUnrKtSrlAfhmi5FBdlP6yrfK5Rn4ZouRQW5Qes61yuUh+GaLkUHsw32W2VSYXyS9jMJ9D7eTALrOtMrlIfhkpLDl8yfUqptNsqywq5Zct387NMEvPLyPXKwuDpMhPdJptlcUl8stIisyk9onrM9sqjyvkl3U05lPQkSgGafbf6DDbKpvR88tIsy+g+PkomkP0lW1Vwsj5ZSONT0R6E2WwfivkUnk8WoHORJGlSR7sJEIao7UmIj2KIsJTga/MoI/Q7lIUEeE58xeE58xXIgYvvlGgNdgS/djvZdeiAPQCogAoQBQABYgCoABRABQgCoACRAFQgCgAChBFzW/6ZZ/bw6mvJoiiJPqfz4H38uGtZ6ivLoiSQe+7X6nPHkTZoowE1YYnqA5ZIRa1ZX2aD4dZ1rflHiHdzVFGglrmj9WIRW1YnyqCtlZ+2xCPz74a7yJBgzOMJK0Qi9q0PkUErWl9JbWmgygHHPWht826SdNuHIvavr7PEbTm9RXVmg6i7EiJBC2i41jUvPqkXQRtl08FvjKJkaBF9BqLmlvfrs7cuq7z+OzLkh4J+vIXUt7kirGoTepbXtAmgrbHx2dfk5xI0NJD9haLWlifNIygZYxyDnmRoKX0FYtapb6MVfgLPT77YuRGglodu4dYVJP6GkfQ9vn47ItQEAlqV0LjWFTT+tpG0LIyD9AYRAFQgCgAChAFQAGiAChAFAAFiAKgAFEAFCAKgAJEAVCAKAAKEAVAAaIAKEAUOA9lblrhQdZPWX76efrOZkSBk1Dmplkc560ov/nHiAL98y43zeaPf2lR8j79iChwOnWzhxEFLkCN3LTUz9kjCvSNIjctL1zi6SC0KDAy5blpOhAFhuXM3DREgUFJyU2j6wX3pGVu2tuSWEeBnjgpN03fCm0WHKe0HQKIAqAAUQAUIAqAAkQBUIAoAAoQBUABogAoQBQABYgCoABRABQgCoACRAFQgCgACv4HiaGdOUOCt2UAAAAASUVORK5CYII=[/img][/center] Change the value of slider [i][color=#0000ff]λ[/color][/i]. Only two values induce singular matrix [color=#0000ff](M-λE)[/color].[list][*] [i][color=#0000ff]λ=1[/color][/i] gives direction vector of mirror line [color=#0000ff](M-E)X = o; x + 2y = 0, X1=(-2t, t)[/color][/*][*] [i][color=#0000ff]λ=-1[/color][/i] gives f normal vector of mirror line [color=#0000ff](M+E)X = o; x - 0.5y = 0, X2 = (t, 2t) [/color][br][/*][/list]