Question

kite efgh is the image of kite efgh under a translation right 8 units and down 4 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

For a translation right \(8\) units and down \(4\) units, the rule for translation in the coordinate plane is \((x,y)\to(x + 8,y-4)\). In general, for a horizontal translation \(a\) units (right if \(a>0\), left if \(a < 0\)) and vertical translation \(b\) units (up if \(b>0\), down if \(b < 0\)), the rule is \((x,y)\to(x + a,y + b)\). Here \(a = 8\) and \(b=-4\).

Step2: Determine the reflection rule

Let's assume we first do the translation \((x,y)\to(x + 8,y-4)\). Let's take a point, say \(E(-6,-1)\). After translation \(E\to E_1(-6 + 8,-1-4)=(2,-5)\). The image of \(E\) after the two - step transformation (translation + reflection) is \(E'(1,5)\).
We know that the reflection over the \(y\) - axis has the rule \((x,y)\to(-x,y)\). Let's check with another point. Take \(F(-5,2)\). After translation \(F\to F_1(-5 + 8,2-4)=(3,-2)\). If we then reflect \((3,-2)\) over the \(y\) - axis, we get \((-3,-2)\), but we want \(F'(2,2)\). Wait, no. Let's use the fact that if we consider the composition.
Let's assume the translation \((x,y)\to(x + 8,y-4)\) followed by reflection. Let's use the origin - based approach.
The mid - point between a point \(P(x,y)\) and its image \(P'(x',y')\) after reflection over a line can help. But another way:
We know that if we first translate \((x,y)\to(x + 8,y-4)\) and then reflect. Let's assume the reflection is over the \(y\) - axis.
Let \(x_1=x + 8\) and \(y_1=y-4\). After reflection over the \(y\) - axis, the rule for reflection over the \(y\) - axis is \((x_1,y_1)\to(-x_1,y_1)\). Substituting \(x_1=x + 8\) and \(y_1=y-4\) back, we get \((x,y)\to(-(x + 8),y-4)\). But let's check with \(E(-6,-1)\):
After translation \(x=-6+8 = 2,y=-1 - 4=-5\), then reflection over \(y\) - axis \(x=-2,y=-5\) (wrong).
Let's check the \(x\) - coordinates. The \(x\) - coordinate of \(E\) is \(-6\), and of \(E'\) is \(1\). The \(x\) - coordinate of \(F\) is \(-5\) and of \(F'\) is \(2\). The difference between \(x\) (original) and \(x'\) (final) for \(E\): \(-6\to1\), for \(F\): \(-5\to2\). The pattern is \(x\to - x-5\) (no). Wait, using the translation first \((x,y)\to(x + 8,y-4)\) and then assume reflection over the line \(x = 3\). The rule for reflection over the line \(x=a\) is \((x,y)\to(2a - x,y)\). If \(a = 3\), then \((x_1,y_1)\to(6 - x_1,y_1)\). Substituting \(x_1=x + 8\) and \(y_1=y-4\), we get \((x,y)\to(6-(x + 8),y-4)=(-x - 2,y-4)\) (wrong).
Let's use a better approach.
We know that the transformation is a composition of translation and reflection.
Take a general point \((x,y)\). After translation \((x,y)\to(x + 8,y-4)\). Let's assume the reflection is over the \(y\) - axis. But no. Let's use vectors.
Another way:
We can also use the fact that if we consider the transformation of the \(x\) - coordinate.
Let's take \(E(-6,-1)\). After translation right \(8\) units: \(x=-6 + 8=2\), \(y=-1-4=-5\). The final \(E'\) has \(x = 1\), \(y = 5\).
Wait, no. Let's use the formula for translation \((x,y)\to(x + 8,y-4)\) and then reflection over the \(x\) - axis. The rule for reflection over the \(x\) - axis is \((x,y)\to(x,-y)\). If we first translate \(E(-6,-1)\) to \((2,-5)\) and then reflect over \(x\) - axis \((2,5)\) (wrong \(x\) - coordinate).
Let's use the fact that if we assume the reflection is over the line \(x=\frac{1}{2}\). No.
Let's use the following:
We know that the transformation is a glide - reflection (translation + reflection).
The translation rule is \((x,y)\to(x + 8,y-4)\).
Let's assume the reflection is over the \(y\) - axis. But we can also check the symmetry.
Let's use the fact that if we consider the composition of translation \(\vec{T}=(8,-4)\) and reflection.
Let's take a point \(G(-2,1)\). After translation \(G\to G_1(-2 + 8,1-4)=(6,-3)\). The image \(G'\) is \((6,3)\). The reflection over the \(x\) - axis: \((x,y)\to(x,-y)\). If we first translate \((x,y)\to(x + 8,y-4)\) and then reflect over the \(x\) - axis \((x + 8,y-4)\to(x + 8,-(y - 4))=(x + 8,-y + 4)\).

Answer:

Translation: \((x,y)\to(x + 8,y-4)\)
Reflection: \((x,y)\to(x,-y)\)