Final answer:
The coordinates for the final figure after rotating 90° clockwise about the origin and dilating with respect to the origin using a scale factor of 3 are J''(-3,3), K''(-12,9), L''(-3,15).
Step-by-step explanation:
To find the coordinates of the figure after the given transformations, we first need to rotate the figure 90° clockwise about the origin. To do this, we can use the rotation matrix:
$$ \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $$
Since we are rotating 90° clockwise, our angle of rotation will be -90°. Plugging in the coordinates of each vertex of the figure into the rotation matrix, we get:
$$ \begin{align*} J' &= (1,1) \cdot \begin{bmatrix} \cos(-90°) & -\sin(-90°) \\ \sin(-90°) & \cos(-90°) \end{bmatrix} = (1,1) \cdot \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = (-1,1) \\ K' &= (3,4) \cdot \begin{bmatrix} \cos(-90°) & -\sin(-90°) \\ \sin(-90°) & \cos(-90°) \end{bmatrix} = (3,4) \cdot \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = (-4,3) \\ L' &= (5,1) \cdot \begin{bmatrix} \cos(-90°) & -\sin(-90°) \\ \sin(-90°) & \cos(-90°) \end{bmatrix} = (5,1) \cdot \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = (-1,5) \end{align*} $$
After rotating the figure, we need to dilate it with respect to the origin using a scale factor of 3. To do this, we simply multiply the coordinates of each vertex by the scale factor:
$$ \begin{align*} J'' &= (-1,1) \cdot 3 = (-3,3) \\ K'' &= (-4,3) \cdot 3 = (-12,9) \\ L'' &= (-1,5) \cdot 3 = (-3,15) \end{align*} $$
Therefore, the coordinates for the final figure after the given transformations are:
$$ J''(-3,3), K''(-12,9), L''(-3,15) $$