Question

30 POINTS AND BRAINLIEST

Answer 1

We could find the rotation point using matrix $\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right]$

X(-4,7) is rotated to (0, 90°)
X' = $\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right] \left[\begin{array}{ccc}-4\\7\end{array}\right]$
X' = $\left[\begin{array}{ccc}(0\times-4)-1\times7\\(1\times-4)-0\times7\end{array}\right]$
X' = $\left[\begin{array}{ccc}-7\\-4\end{array}\right]$
X' = (-7,-4)

Y(6,2) is reflected to (0,90°)
Y' = $\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right] \left[\begin{array}{ccc}6\\2\end{array}\right]$
Y' = $\left[\begin{array}{ccc}(0\times6)-1\times2\\(1\times6)-0\times2\end{array}\right]$
Y' = $\left[\begin{array}{ccc}-2\\6\end{array}\right]$
Y' = (-2,6)

Z(3,-8) is reflected to (0,90°)
Z' = $\left[\begin{array}{ccc}0&-1\\1&0\end{array}\right] \left[\begin{array}{ccc}3\\-8\end{array}\right]$
Z' = $\left[\begin{array}{ccc}(0\times3)-1\times-8\\(1\times3)-0\times-8\end{array}\right]$
Z' = $\left[\begin{array}{ccc}8\\3\end{array}\right]$
Z' = (8,3)

See image attached