1. If [3, -5] is the vector for a translation, what is the image of the point (0,0)?

Ans: (3, -5)

3.If [3, -5] is the vector for a translation, what is the image of the point (-1, -8)?

Ans:(2, -13)

5. If [5,9] is the vector for a translation, what point has the image (0,0)?

Ans: (-5, -9).

7. If the second point is the image of the first under a translation, give the vector for the translation.

1. (3, 4) (5, -2)
2. (-1, -6) ((-3, -1)

Ans:

1. [2, -6]
2. [-2, 5]

9. Find the image of the line y = 7x under the translation with vector [3, 4/5].

Ans: 7x' - y' - 101/5 = 0. Solution: The translation is given by x' = x + 3, y' = y + 4/5. Therefore, x = x' - 3 and y = y' - 4/5. Substituting these in the equation of the line gives, y' - 4/5 = 7(x' - 3).

11. What is the image of 3x + 4y + 2 = 0 under the translation with vector [4, -3]?

Ans:3x' + 4y' + 2 = 0. Solution: The translation is given by x' = x + 4, y' = y - 3. Solve these for x and y and substitute the values into the original equation.

13. What is the image of the point (3,4) under a rotation of 45º about the origin?

Ans: (- sqrt(2)/2, 7 sqrt(2)/2). Solution: For a rotation of 45º about the origin, the transformation is given by x' = x cos 45º - y sin 45º, y' = x sin 45º + y cos 45º. Since sin 45º = cos 45º = sqrt(2)/2, we get x' = 3sqrt(2)/2 - 4sqrt(2)/2 and y' = 3sqrt(2)/2 + 4sqrt(2)/2.

15. What point has the image (-1, 2) under a rotation of 45º about the origin?

Ans: (sqrt(2)/2, 3 sqrt(2)/2). Solution: For a rotation of 45º about the origin, the transformation is given by x' = x cos 45º - y sin 45º, y' = x sin 45º + y cos 45º. Since sin 45º = cos 45º = sqrt(2)/2, we get x' = xsqrt(2)/2 - ysqrt(2)/2 and y' = xsqrt(2)/2 + ysqrt(2)/2. Given that x' = -1 and y' = 2, we solve this system of two equations and two unknowns for x and y.

17. Derive the equations for the inverse of a transformation of rotation about a point (h,k), with angle of rotation \theta.

Solution: x - h = (x' - h) cos \theta + (y' - k) sin \theta, y- k = (x' - h) sin \theta + (y' - k) cos \theta.

19. Find the image of (3,7) under a reflection about the line with \theta = 60º, \phi = 330º, and d = 5.

Ans: ((17 sqrt(3) - 3)/2, (3 sqrt(3) - 3)/2). Solution: The formula for a general reflection is given by x' = x cos 2(\theta) + y sin 2(\theta) + 2d cos (\phi), y' = x sin 2(\theta) - y cos 2(\theta) + 2d sin (\phi). With the values given, we have x' = 3 cos 120º + 7 sin 120º + 10 cos 330º, y' = 3 sin 120º - 7 cos 120º + 10 sin 330º.

21. Find the inverse transformation for the transformation whose equations are x' = x - 5, y' = y + 2.

Solution: x = x' + 5, y = y' - 2.

23. Find the inverse transformation for the transformation whose equations are x' = -x, y' = -y.

Solution: x = -x', y = -y'

25. Find the inverse transformation for the transformation whose equations are x' = x cos 30º - y sin 30º, y' = x sin 30º + y cos 30º. .

Solution: x = x' cos 30º + y' sin 30º, y = -x' sin 30º + y' cos 30º.

27. Derive the set of equations for the product of two translations whose vectors are [6, 4], [-2,3].

Solution: x' = x + 4, y' = y + 7

29. Derive the general formula for the image of a point under the product of the two translations with vectors [a,b] and [c,d].

Solution: x" = x + (a + c), y" = y + (b + d).