Question

triangle abc is the image of triangle abc under a translation right 4 units and down 5 units followed by a reflection.
write the rules for the translation and reflection.
translation: (x, y) ↦ (□, □)
reflection: (x, y) ↦ (□, □)

Explanation:

Step1: Determine the translation rule

A translation right \(4\) units means we add \(4\) to the \(x -\)coordinate, and a translation down \(5\) units means we subtract \(5\) from the \(y -\)coordinate.
The rule for translation is \((x,y)\to(x + 4,y-5)\).

Step2: Determine the reflection rule

Let's assume after translation, we reflect over the \(y -\)axis. The rule for reflection over the \(y -\)axis is \((x,y)\to(-x,y)\). We can check with a point. For example, take point \(A(1,3)\). After translation \(A_1=(1 + 4,3-5)=(5,-2)\). If we assume reflection over the \(y -\)axis, then \(A'=(-5,-2)\) (but wait, no. Wait, let's check the general form.

Another way: Let's use the fact that if we first translate \((x,y)\to(x + 4,y - 5)\) and then reflect. Let's take two - step transformation. Suppose we consider the composition. Let's check the mid - point of \(A\) and \(A'\). The coordinates of \(A(1,3)\) and \(A'(5,2)\) (wait no, wait \(A(1,3)\): after translation \((1 + 4,3-5)=(5,-2)\), but \(A'\) is \((5,2)\). So after translation \((x,y)\to(x + 4,y-5)\), then reflection over the \(x -\)axis \((x,y)\to(x,-y)\).
Check: For point \(A(1,3)\), translation: \((1+4,3 - 5)=(5,-2)\), reflection over \(x -\)axis \((5,2)\) (matches \(A'\)). For point \(C(-2,-2)\), translation \((-2 + 4,-2-5)=(2,-7)\), reflection over \(x -\)axis \((2,7)\) (matches \(C'\)).

Answer:

Translation: \((x,y)\to(x + 4,y-5)\)
Reflection: \((x,y)\to(x,-y)\)