A triangle has vertices A(0,0) B(4,0) and C(2,3). It is translated by 4 units in X-direction and 2 Y units in Y-direction. It is then rotated by $90^0$ in anticlockwise direction about the new position of point C. Find the new vertices of a triangle.

Mumbai University > Mechanical Engineering > Sem 7 > CAD CAM CAE

Marks: Marks

Year: Dec 2016

A triangle PQR has its vertices at P(0,0), Q(4,0), and R(2,3) . It is to be translate by 4 units in X-direction, and 2 units in Y-direction, then it is to be rotated in anticlockwise direction about the new position of point R through 90 Degree. Find the new position of the triangle.

Refer Fig 1.12.8.1

$\triangle PQR = \begin{bmatrix} 0&0&1 \\ 4&0&1 \\ 2&3&1 \end{bmatrix}$

a) To be translated by $t_x=4$ and $t_y=2$

$\triangle P'Q'R'$ $= \triangle PQR.T \\ =\begin{bmatrix} 0&0&1 \\ 4&0&1 \\ 2&3&1 \end{bmatrix} \begin{bmatrix}1&0&0 \\ 0&1&0 \\ 4&2&1 \end{bmatrix} \\ =\begin{bmatrix} 4&2&1 \\8&2&1 \\ 6&5&1 \end{bmatrix}$

Besides a rotary arrangement, a twin table arrangement can also be used for work loading as shown in Linear Twin Pallet System . the working of this system is similar to that of a rotary pallet system. However, the work tables in this case are not indexed in rotary form.

Another form of a work changer is the one having a conveying system to continuously feed the machine with the work-piece. In places where work loading is extremely sensitive, robots can also be used for automated work changing.

b) To be rotated in anticlockwise direction about the new point R' (6,5) through $90^0$

$\triangle P''Q''R''$ $=\triangle P'Q'R'.T.R.T^{-1} \\ = \begin{bmatrix} 4&2&1 \\8&2&1 \\ 6&5&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ -6&-5&1 \end{bmatrix} \begin{bmatrix} \cos 90 & \sin 90 & 0 \\ -\sin 90 & \cos 90 & 0 \\ 0&0&1\end{bmatrix}\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 6&5&1 \end{bmatrix} \\ =\begin{bmatrix} 4&2&1 \\ 8&2&1 \\ 6&5&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ -6&-5&1 \end{bmatrix} \begin{bmatrix}0&1&0 \\ -1&0&0 \\ 0&0&1 \end{bmatrix}\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 6&5&1 \end{bmatrix} \\ =\begin{bmatrix} 4&2&1 \\ 8&2&1 \\ 6&5&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ -6&-5&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 6&5&1 \end{bmatrix} \\ =\begin{bmatrix} 4&2&1 \\8&2&1 \\ 6&5&1 \end{bmatrix} \begin{bmatrix} 0&1&0 \\ -1&0&0 \\ 11&-1&1\end{bmatrix} \\ = \begin{bmatrix}9&3&1 \\ 9&7&1 \\ 6&5&1\end{bmatrix}$

New vertices are P''(9,3) Q''(9,7) R''(6,5)