Question

A scaling factor of 2 is applied in the y direction while no scaling is applied in the x direction to the line whose two end points are at coordinates (1, 3) and (3, 6). The line is to be rotated subsequently through 30 in the counter-clockwise direction. Determine the necessary transformation matrix for the operation and the new coordinates of the end points.

Answer 1

Step-by-step explanation:

First, scaling by a factor of 2 in the y direction while not scaling in the x direction (or we could say "scaling by a factor of 1"), we're multiplying (on the left) by the matrix

$\left(\begin{array}{cc}1&0\\0&2\end{array}\right)$

Next, to rotate counter-clockwise by an angle of θ, the relevant matrix is

$\left(\begin{array}{cc}\cos\theta &-\sin\theta\\ \sin\theta & \cos\theta \end{array}\right)$

For θ = 30 degrees, we have cos θ = √3 / 2 and sin θ = 1 / 2, so the rotation matrix is

$\left(\begin{array}{cc}\sqrt{3}/2 & -1/2 \\ 1/2 & \sqrt{3}/2\end{array}\right)$

The result of doing these two operations in succession is (remember we're operating on the left, so the last one to be done is the left-most one in the product):

$\left(\begin{array}{cc}\sqrt{3}/2 & -1/2 \\ 1/2 & \sqrt{3}/2\end{array}\right) \left(\begin{array}{cc}1&0\\0&2\end{array}\right) = \left(\begin{array}{cc}\sqrt{3}/2 & -1 \\ 1/2 & \sqrt{3}\end{array}\right)$

To find the new coordinates of the given points, we just multiply the given points by this matrix.  Thus (1, 3) goes to

( √3 / 2 - 3,  1/2 + 3√3 )

and (3, 6) goes to

( 3√3 / 2 - 6,  3/2 + 6√3 )