Question

mica translated polygon c two units to the right and then reflected the image across the y-axis to get polygon t. sasha used one transformation to transform polygon c to polygon t. describe the transformation that sasha used.

complete the sentence:
this is a reflection across the ______

Explanation:

Step1: Analyze Mica's Transformations

Mica first translates Polygon C 2 units right. Let a point \((x,y)\) on C become \((x + 2,y)\) after translation. Then reflecting over y - axis: a point \((x,y)\) reflects to \((-x,y)\), so \((x + 2,y)\) becomes \((-(x + 2),y)=(-x - 2,y)\).

Step2: Find Sasha's Single Transformation

We want a single transformation from \((x,y)\) on C to \((-x - 2,y)\) on T. Let's consider reflection over a vertical line \(x = a\). The reflection of \((x,y)\) over \(x=a\) is \((2a - x,y)\). We set \(2a - x=-x - 2\), solving for \(a\): \(2a=-2\), so \(a = - 1\). Wait, alternatively, notice the horizontal distance between C and T. Looking at the graph, Polygon C is on the left, T on the right. Wait, no, let's think about the effect of translation then reflection. Translating 2 right then reflecting over y - axis is equivalent to reflecting over the line \(x=-1\)? Wait, no, let's take a point. Suppose a vertex of C is at \(x=-3\), \(y = 2\). Translate 2 right: \(x=-3 + 2=-1\), \(y = 2\). Reflect over y - axis: \(x = 1\), \(y = 2\). Now, what's the reflection of the original point \((-3,2)\) to \((1,2)\)? The midpoint between \(-3\) and \(1\) is \(\frac{-3 + 1}{2}=-1\). So the reflection line is \(x=-1\)? Wait, no, wait the problem's graph: Polygon C is left of y - axis, T is right. Wait, maybe I made a mistake. Wait, Mica translates 2 right, then reflects over y - axis. Let's take a general point \((x,y)\) in C. After translating 2 right: \((x + 2,y)\). Reflect over y - axis: \((- (x + 2),y)=(-x - 2,y)\). Now, we want a single transformation from \((x,y)\) to \((-x - 2,y)\). Let's consider reflection over the line \(x=-1\). The reflection of \((x,y)\) over \(x = - 1\) is \((-2 - x,y)\) (since the formula for reflection over \(x=a\) is \(x'=2a - x\), here \(a=-1\), so \(x'=2(-1)-x=-2 - x\)), which is the same as \(-x - 2\). Alternatively, looking at the horizontal shift: the net effect of translating 2 right and reflecting over y - axis is a reflection over the line \(x=-1\)? Wait, no, maybe a reflection over a vertical line. Wait, another approach: the distance between the original and the final. The translation 2 right and reflection over y - axis: let's see the horizontal change. Original x, after translation x + 2, after reflection -x - 2. The difference between -x - 2 and x is -2x - 2, but that's not helpful. Wait, looking at the graph, Polygon C and T are symmetric with respect to the line \(x=-1\)? Wait, no, maybe the line \(x=-1\) is the vertical line midway between the original and the transformed. Wait, maybe I messed up. Wait, let's take the first part: Mica's steps. Translate 2 right: so each point moves 2 units in x - direction. Then reflect over y - axis: each point \((x,y)\) becomes \((-x,y)\). So combining, \((x,y)\to(x + 2,y)\to(-(x + 2),y)=(-x - 2,y)\). Now, Sasha does it in one step. Let's find the reflection line. Suppose we reflect over \(x = a\). Then the reflection of \((x,y)\) is \((2a - x,y)\). We want \(2a - x=-x - 2\), so \(2a=-2\), \(a=-1\). So the transformation is a reflection across the vertical line \(x=-1\). Wait, but looking at the graph, Polygon C is on the left, T on the right. Wait, maybe the line \(x=-1\) is between them. Alternatively, maybe the reflection is over the line \(x=-1\). Let's check with a point. Suppose a vertex of C is at \(x=-4\), \(y = 2\). Translate 2 right: \(x=-2\), \(y = 2\). Reflect over y - axis: \(x = 2\), \(y = 2\). Now, reflect \((-4,2)\) over \(x=-1\): \(2(-1)-(-4)=-2 + 4 = 2\), which matches. So yes, the single transformation is a reflection across the vertical line \(x=-1\).

Answer:

Sasha used a reflection across the vertical line \(x = - 1\) (or the line \(x=-1\)) to transform Polygon C to Polygon T.