Question

question
what is the image point of $(-8,2)$ after the transformation $r_{y = x}circ t_{-3,3}$?
answer attempt 1 out of 2

Explanation:

Step1: Translation

First, perform the translation \(T_{-3,3}\). The rule for translation \(T_{a,b}\) is \((x,y)\to(x + a,y + b)\). For \(a=-3\) and \(b = 3\) and the point \((-8,2)\), we have:
\((x,y)=(-8,2)\)
\(x'=-8+( - 3)=-11\)
\(y'=2 + 3=5\)
The point after translation is \((-11,5)\)

Step2: Reflection over \(y = x\)

The rule for reflection over the line \(y=x\) is \((x,y)\to(y,x)\). For the point \((-11,5)\) after translation, when we apply the reflection \(r_{y = x}\), we swap the \(x\) and \(y\) - coordinates.
If \((x,y)=(-11,5)\), then the image point after reflection is \((5,-11)\)

Answer:

\((5,-11)\)