I.G..S.E. Matrices and Transformations Index: Please click on the question number you want Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 You can access the solutions from the end of each question
Question 1 Draw x- and y- axis with values from 10 to 10. Draw the following triangle with vertices A(3, 2), B(5, 2) and (5, 4). Draw the image of AB under the following transformations clearly labelling the vertices in each case. Write down the coordinates of the vertices in each case. a. Enlargement scale factor 2, centre (0, 0). Label the image A1, B1, 1. b. Reflection in the y-axis. Label the image A2, B2, 2. c. Rotation 270! about (0, 0). Label the image A3, B3, 3. d. Enlargement scale factor 1, centre (0, 0). Label the image A4, B4, 4. e. 12 Translation 12. Label the image A5, B5, 5. f. Reflection in the line y = x. Label the image A6, B6, 6. g. Enlargement scale factor 2, centre (0, 3). Label the image A7, B7, 7. h. Rotation 90! about the origin. Label the image A8, B8, 8. lick here to read the solution to this question lick here to return to the index
Solution to question 1 y 10 1 B 7 A 7 8 B 8 5 2 A 1 B 1 A 8 B 2 A 2 A B 7-10 -5 5 10 x B 4 A 4 A 6 A 3 4-5 6 B 6 B 3 3 5 A 5 B 5 lick here to return to question 2-10 y = x a. A1, B1, 1is an enlargement scale factor 2, centre (0, 0). A1(6,1), B1(10, 4), 1(10, 8). b. A2, B2, 2is a reflection in the y-axis. A2 ( 3,2), B2 ( 5,2), 2 ( 5,4). c. A3, B3, 3 is a rotation 270! (anticlockwise) about (0, 0). A (2, 3), B (2, 5), (4, 5). 3 3 3 lick here to continue with solution or go to next page
d. A4, B4, 4 is an enlargement scale factor 1, centre (0, 0). A ( 3, 2), B ( 5, 2), ( 5, 4). 4 4 4 e. A5, 5, 5 B is a translation 12 12. A5 ( 9, 10), B5 ( 7, 10), 5 ( 7, 8). f. A6, B6, 6 is a reflection in the line y = x. A ( 2, 3), B ( 2, 5), ( 4, 5). 6 6 6 g. A7, B7, 7is an enlargement scale factor 2, centre (0, 3). A ( 6,5), B ( 10,5), ( 10,1). 7 7 7 h. A8, B8, 8 is a rotation 90! about the origin. A ( 2,3), B ( 2,5), ( 4,5) 8 8 8 lick here to read the question again lick here to return to the index
Question 2 From the diagram in question 1 describe the single transformation, which maps a. A2, B2, 2 to A3, B3, 3 b. A4, B4, 4 to A6, B6, 6 c. A8, B8, 8 to A2, B2, 2 d. A3, B3, 3 to A4, B4, 4 e. A6, B6, 6 to A3, B3, 3 lick here to read the solution to this question lick here to return to the index
Solution to question 2 lick here to see the diagram a. A2, B2, 2 to A3, B3, 3 is a reflection in the line y = x. b. A4, B4, 4 to A6, B6, 6 is a reflection in the line y = x. c. A8, B8, 8 to A2, B2, 2 is a reflection in the line y = x. d. A3, B3, 3 to A4, B4, 4 is a rotation 270! about (0, 0). e. A6, B6, 6 to A3, B3, 3 is a reflection in the y-axis or x = 0. lick here to read the question again lick here to return to the index
Question 3 Draw x- and y- axis with values from 10 to 10. Draw the following triangle with vertices A(5, 4), B(9, 4) and (5, 6) and its image under a rotation A 1(-9, -3), B 1 (-9, -7) and 1(-7, -3). Show by construction the centre of rotation. lick here to read the solution to this question lick here to return to the index
Solution to question 3 y entre of rotation A B Perpendicular bisector of AA 1 x A 1 1 Perpendicular bisector of BB 1 B 1 The centre of rotation is found by joining two corresponding vertices A to A 1 and B to B 2. Then construct the perpendicular bisectors of both lines using a compass and a ruler. Finally the point of intersection of the two perpendicular bisectors is the centre of rotation. lick here to read the question again lick here to return to the index
Question 4 A is a rotation 270! about (0, 0) B is a reflection in the line y = 2 is a translation, which maps (-2, 3) to (2, 4) Find the image of the point (-3, 2) under the following transformations a. A b. A 2 c. B d. AB e. B -1-1 lick here to read the solution to this question lick here to return to the index
Solution to question 4 A is a rotation 270! about (0, 0) B is a reflection in the line y = 2 is a translation, which maps (-2, 3) to (2, 4). Note: this is a translation 4 1 y (3, 2) A ( -1 ) (AB) (A 2 ) x y = 2 (B -1-1 ) (B) B (B) a. A (2, 3). b. A 2 is the same as a rotation of 180!. (3, -2). c. B is a reflection in the line y = 2 followed by a translation 4. (1, -5) 1 4 d. AB is a translation followed by a reflection in the line 2 1 y = followed by a rotation of 270!. (-7, -1) e. B -1-1 4 are inverse transformations. This is a translation followed by 1 a reflection in the line y = 2 (self inverse). (-7, -5). lick here to read the question again lick here to return to the index
Question 5 The transformation T is given by transformations. ' x 2 0 4 ' = + y 0 2 3 is composed of two a. Describe the two transformations. b. Find the image of the point (2, -1) under the transformation. c. Find the point, which is mapped by T onto the point (6, 7). lick here to read the solution to this question lick here to return to the index
Solution to question 5 T is given by 2 0 4 ' = + y 0 2 3 ' x a. ' x 2 0 4 ' = + is an enlargement scale factor 2 centre (0, 0), y 0 2 3 4 followed by a translation 3. b. The image of (2, -1) is given by ' x 2 0 2 4 ' = + y 0 2 1 3 4+ 0 4 = + 0+ 2 3 4 4 = + 2 3 0 = 1 which is (0, -1). c. To find the point, which is mapped by T onto the point (6, 7), we must work with the inverse transformations. 4 First the point (6, 7) is mapped to (2, 10) by the translation 3. The inverse of an enlargement scale factor 2 is an enlargement scale 1 1 0 factor, given by the matrix 2 2 1 0 2 1 0 x 2 2 1+ 0 1 = = =, which gives the point (-1, -5). y 1 10 0 5 5 0 2 lick here to read the question again lick here to return to the index
Question 6 A is a reflection in the line y = x. B is a reflection in the x-axis. Find the matrix, which represents a. A b. B c. AB d. BA Describe the single transformations AB and BA. lick here to read the solution to this question lick here to return to the index
Solution to question 6 A is a reflection in the line y = x. B is a reflection in the x-axis. a. A b. B J (-1, 0) J(0, 1) y 1 I(1, 0) I (0, -1) y = x x 0 0 1 0 1 1 0 0 1 A = 1 0 J(0, 1) y I (1, 0) I(1, 0) x 1 1 0 0 0 0 1 1 J (0, -1) 1 0 B = 0 1 0 1 1 0 0 1 c. AB = = 1 0 0 1 1 0 From the diagram we can see that AB is a rotation 270! about (0, 0). y J(0, 1) J (-1, 0) I(1, 0) x I (0, -1) y J(0, 1) I (0, -1) 1 0 0 1 0 1 d. BA = = 0 1 1 0 1 0 From the diagram we can see that AB is a rotation 90! about (0, 0). J (-1, 0) I(1, 0) x lick here to read the question again lick here to return to the index