 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
SYMMETRY in Visual Studio .NET
SYMMETRY Scan QRCode In .NET Framework Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. Generating Denso QR Bar Code In .NET Framework Using Barcode generation for .NET framework Control to generate, create QR Code ISO/IEC18004 image in .NET framework applications. [CHAP. 6
Read QR Code JIS X 0510 In VS .NET Using Barcode reader for .NET Control to read, scan read, scan image in .NET framework applications. Bar Code Creator In VS .NET Using Barcode encoder for .NET framework Control to generate, create barcode image in .NET framework applications. Fig. 65 EXAMPLES
Reading Bar Code In .NET Framework Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET framework applications. Creating Quick Response Code In C# Using Barcode generator for .NET framework Control to generate, create QR Code JIS X 0510 image in .NET framework applications. (a) The ellipse graphed in Fig. 64(b) is symmetric with respect to the origin because x2 + y2 = 1 4 ( x)2 + ( y)2 = 1 4 Create Denso QR Bar Code In Visual Studio .NET Using Barcode generation for ASP.NET Control to generate, create QR image in ASP.NET applications. Creating QR Code ISO/IEC18004 In Visual Basic .NET Using Barcode creation for Visual Studio .NET Control to generate, create QR image in .NET applications. implies
Make GS1 DataBar Limited In VS .NET Using Barcode maker for VS .NET Control to generate, create GS1 DataBar Expanded image in .NET applications. Code 128 Code Set C Creator In Visual Studio .NET Using Barcode encoder for VS .NET Control to generate, create Code 128 Code Set B image in .NET applications. (b) The hyperbola xy = 1 (see Fig. 66) is symmetric with respect to the origin, for if xy = 1, then ( x)( y) = 1. (c) If y = ax, then y = a( x). Hence, any straight line through the origin is symmetric with respect to the origin. Creating GTIN  12 In VS .NET Using Barcode encoder for Visual Studio .NET Control to generate, create UCC  12 image in .NET framework applications. Printing ISSN In Visual Studio .NET Using Barcode maker for .NET framework Control to generate, create ISSN image in .NET applications. Fig. 66 DataMatrix Recognizer In C#.NET Using Barcode reader for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications. Bar Code Generation In None Using Barcode creator for Software Control to generate, create bar code image in Software applications. Solved Problems
Print Data Matrix 2d Barcode In Java Using Barcode creation for Java Control to generate, create ECC200 image in Java applications. Painting Barcode In ObjectiveC Using Barcode creator for iPhone Control to generate, create barcode image in iPhone applications. 6.1 Determine whether the line y = x (see Fig. 67) is symmetric with respect to: (c) the origin. (a) the xaxis; (b) the yaxis; EAN / UCC  13 Encoder In Java Using Barcode maker for Java Control to generate, create EAN13 image in Java applications. Print ECC200 In None Using Barcode maker for Excel Control to generate, create Data Matrix image in Office Excel applications. (a) The line is not symmetric with respect to the xaxis, since ( 1, 1) is on the line, but ( 1, 1), the re ection of ( 1, 1) in the xaxis, is not on the line. (b) The line is not symmetric with respect to the yaxis, since ( 1, 1) is on the line, but (1, 1), the re ection of ( 1, 1) in the yaxis, is not on the line. (c) The line is symmetric with respect to the origin, by example (c) above. Bar Code Printer In None Using Barcode printer for Microsoft Excel Control to generate, create bar code image in Microsoft Excel applications. ANSI/AIM Code 128 Encoder In None Using Barcode drawer for Software Control to generate, create Code 128 Code Set A image in Software applications. CHAP. 6] SYMMETRY
Fig. 67 Fig. 68 6.2 Determine whether the parabola x = y2 (see Fig. 68) is symmetric with respect to: (c) the origin. (a) The parabola is symmetric with respect to the xaxis, since x = y2 implies x = ( y)2 . (a) the xaxis; (b) the yaxis; (b) It is not symmetric with respect to the yaxis, since (1, 1) is on the parabola, but ( 1, 1) is not. (c) It is not symmetric with respect to the origin, since (1, 1) is on the parabola, but ( 1, 1) is not. 6.3 Show that if the graph of an equation f (x, y) = 0 is symmetric with respect to both the xaxis and the yaxis, then it is symmetric with respect to the origin. (However, the converse is false, as is shown by Problem 6.1.) Assume that f (x, y) = 0; we must prove that f ( x, y) = 0. Since f (x, y) = 0 and the graph is symmetric with respect to the xaxis, f (x, y) = 0. Then, since f (x, y) = 0 and the graph is symmetric with respect to the yaxis, f ( x, y) = 0. 6.4 Let points P and Q be symmetric with respect to the line L : y = x. If P has coordinates (a, b), show that Q has coordinates (b, a). Let Q have coordinates (u, v), and let B be the intersection point of L and the line PQ (see Fig. 69). B bisects PQ; hence, its coordinates are given by (2.2) as a+u b+v , 2 2 from which, since B lies on L , a+u b+v = 2 2 b+v =a+u v u=a b Furthermore, the perpendicular lines PQ and L have respective slopes b v a u (1) SYMMETRY
[CHAP. 6
so that, by Theorem 4.2, b v (1) = 1 a u b v =u a b+a=v+u v+u=a+b (2) To solve (1) and (2) simultaneously for u and v, rst add the two equations, yielding 2v = 2a, or v = a. Then subtract (1) from (2), yielding 2u = 2b, or u = b. Thus, Q has coordinates (b, a). Fig. 69 6.5 If the graph of 3x 2 + xy = 5 is re ected in the yaxis (that is, each point on the graph is replaced by the point symmetric to it with respect to the yaxis), nd an equation of the new graph. A point (x, y) lies on the new graph if and only if ( x, y) is on the original graph; that is, if and only if 3( x)2 + ( x)y = 5. This is equivalent to 3x 2 xy = 5. Note that the new equation is obtained from the old equation by replacing x by x.

