Graph the reflection of delta ABC in the line n x 1 The imag

Graph the reflection of delta ABC in the line n: x =1 The image can be found by multiplying [-1 0 0 1] and [-1 4 2 -2 -1 1] or using the transformation (x, y) rightarrow (-x, y). T/F y is the perpendicular bisector of BB\'. Use the figure to the right. Use the coordinate rules to rotate delta ABC 90 degree counterclockwise around the origin. List the coordinates and draw the image of delta A\'B\'C\'. (a, b) rightarrow (______, ______) A(-1, -2) rightarrow A\'(______, ______) B(1, 2) rightarrow B\'(_______, ______) C(3, -1) rightarrow C\' (______, ______) Find the image matrix that results from a 270 degree counterclockwise rotation of delta A\'B\'C\" from problem 14. [__ __ __ __] [A\' B\' C\' __ __ __ __ __ __] = [A\" B\" C\" __ __ __ __ __ __] Why you would expect delta A\" B\" C\" to have the same vertices as delta ABC from problem 14?

Solution

Solution --( 11,12 and 13)

11. The reflection line is like a mirror. Point A (- 1, - 2) is 2 units to the left of the line of reflection. Plot A\' 2 units to the right at (3, - 2).

Point B (4, - 1) is 3 units to the right of the line of reflection. Plot B\' 3 units to the left of the line of reflection at (- 3, - 1).

Point C (2, 1) is 1 unit to the right of the line of reflection. Plot C\' 1 unit to the left of the line of reflection at (0, 1).

Notice that for the reflections about a vertical line, a line with equation x = a constant, the y-coordinate of the reflected point does not change. Where the reflection is about a horizontal line, the x-coordinate is the one that doesn\'t change. Where the reflection is about a line of the form y = mx + b, both coordinates of the reflected point will be different.

12. False. The transformation described reflects the triangle about the y-axis.

13. False. The y-axis is perpendicular to segment BB\', but it does not cut it in half. The line of reflection, x = 1 is the perpendicular bisector for BB\' and AA\' and CC\'.