Q25. - Algebra 2

Question

A triangle is rotated $90°$ counter clockwise about the origin. The coordinates of the vertices are $J\text{'}\left(-3,-5\right),K\text{'}\left(-2,7\right),\text{ and }L\text{'}\left(1,4\right)$. What were the coordinates of the triangle in its original position.

Step-by-Step Solution

Answer

Coordinates of triangle in its original position are $J\left(5,-3\right),K\left(-7,-2\right),\text{ and }L\left(-4,1\right)$

Step 1 - Vertex matrix of triangles J K L and J ' K ' L '

Let the coordinates of $J,K,\text{ and }L$ be $\left(x_1,y_1\right),\left(x_2,y_2\right),\text{ and }\left(x_3,y_3\right)$ respectively.

Vertex matrix of coordinates of triangle $JKL$  is $\left[\begin{array}{ccc}{x}_1& x_2& x_3\\ y_1& y_2& y_3\\ & & \end{array}\right]$

Vertex matrix of coordinates of triangle $J\text{'}K\text{'}L\text{'}$ is $\left[\begin{array}{ccc}-3& -2& 1\\ -5& 7& 4\\ & & \end{array}\right]$

Step 2 - Vertex matrix after rotation

Vertex matrix of triangle after rotation of $90°$ counter clockwise is product of matrix $\left[\begin{array}{cc}0& -1\\ 1& 0\\ & \end{array}\right]$ with vertex matrix of given triangle as shown below

$\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\times \left[\begin{array}{ccc}{x}_1& x_2& x_3\\ y_1& y_2& y_3\end{array}\right]=\left[\begin{array}{ccc}{y}_1& y_2& y_3\\ -x_1& -x_2& -x_3\end{array}\right]$

Step 3 - Compare with vertex matrix of triangle J ' K ' L '

Compare vertex matrix of triangle $J\text{'}K\text{'}L\text{'}$ with vertex matrix of triangle after rotation to get the coordinates of triangle in its original position

 $\left[\begin{array}{ccc}{y}_1& y_2& y_3\\ -x_1& -x_2& -x_3\end{array}\right]=\left[\begin{array}{ccc}-3& -2& 1\\ -5& 7& 4\end{array}\right]$ 

 $\begin{array}{c}J\left(x_1,y_1\right)=\left(5,-3\right)\\ K\left(x_2,y_2\right)=\left(-7,-2\right)\\ L\left(x_3,y_3\right)=\left(-4,1\right)\end{array}$