Question

1. triangle abc has vertices a(-2, 5), b(1, 0), and c(6, -2). what are the coordinates of the vertices of △abc for r_{y - axis}?

a a(5, -2), b(0, 1) c(-2, 6)
b a(2, -5), b(-1, 0), c(-6, 2)
c a(2, 5), b(-1, 0), c(-6, -2)
d a(-2, -5), b(1, 0), c(6, 2)

1. what is the rule used to transform △abc to its image?

a(-3, 5), b(2, 8), c(-4, -5) and a(-3, -5), b(2, -8), c(-4, 5)
a r_m(x, y)=(-y, -x), where the equation of line m is y = -x
b r_n(x, y)=(y, x), where the equation of line n is y = -x
c r_{y - axis}(x, y)=(-x, y)
d r_{x - axis}(x, y)=(x, -y)
for items 3 - 5, use △abc.

1. suppose the equation of line s is y = 2. what are the coordinates of c of △abc for r_s?

(, )

1. suppose the equation of line t is y = x. which shows the graph of △abc for r_t?
2. after a reflection of the figure, the images vertices are a(5, 1), b(3, -1), and c(7, -1). what is the line of reflection?

a y = 2
b x = 2
c y = -x
d y = 2x

Explanation:

Step1: Reflection over the y - axis

The rule for reflection over the y - axis is \(r_{y - axis}(x,y)=(-x,y)\).
For point \(A(-2,5)\), applying the rule: \(x=-2\), \(y = 5\), so \(A'=(2,5)\).
For point \(B(1,0)\), applying the rule: \(x = 1\), \(y=0\), so \(B'=(-1,0)\).
For point \(C(6,-2)\), applying the rule: \(x = 6\), \(y=-2\), so \(C'=(-6,-2)\).

Step2: Analyzing the transformation of \(\triangle ABC\) to \(\triangle A'B'C'\)

Given \(A(-3,5)\to A'(-3,-5)\), \(B(2,8)\to B'(2,-8)\), \(C(-4,-5)\to C'(-4,5)\).
The rule for reflection over the x - axis is \(r_{x - axis}(x,y)=(x,-y)\).

Step3: Reflection over the line \(y = 2\)

Let the coordinates of \(C\) be \((x,y)\). From the graph (assuming \(C(-4,0)\) as per the given graph in the problem - like structure).
The distance between \(y\) - coordinate of \(C\) and \(y = 2\) is \(d=|0 - 2|=2\).
The \(y\) - coordinate of \(C'\) is \(y'=2+(2 - 0)=4\) (using the formula for reflection over a horizontal line \(y = k\): if a point \((x,y)\) is reflected over \(y = k\), the new \(y\) - coordinate is \(y'=2k - y\), here \(k = 2\)). The \(x\) - coordinate remains the same. So if \(C(-4,0)\), then \(C'(-4,4)\).

Step4: Reflection over the line \(y=x\)

The rule for reflection over the line \(y = x\) is \(r_{y=x}(x,y)=(y,x)\).
If \(A\) has coordinates (assuming from the graph \(A(-1,2)\)), then \(A'\) has coordinates \((2,-1)\); if \(B\) has coordinates \((2,1)\), then \(B'\) has coordinates \((1,2)\); if \(C\) has coordinates \((-4,0)\), then \(C'\) has coordinates \((0,-4)\).

Step5: Finding the line of reflection

Let's assume the original points (from the graph) \(A(1,5)\), \(B(3,1)\), \(C(7,1)\) (by using the mid - point formula. For a point \(P(x,y)\) and its image \(P'(x',y')\) after reflection over a line \(x = a\), the mid - point of \(PP'\) lies on the line \(x=a\) and the line \(PP'\) is perpendicular to \(x = a\).
The mid - point of \(A(1,5)\) and \(A'(5,1)\) is \((\frac{1 + 5}{2},\frac{5+1}{2})=(3,3)\). The mid - point of \(B(3,1)\) and \(B'(3,-1)\) is \((3,0)\). The mid - point of \(C(7,1)\) and \(C'(7,-1)\) is \((7,0)\).
The line \(x = 2\) is not correct.
For reflection over \(x=2\), if a point \((x,y)\) is reflected over \(x = 2\), the new \(x\) - coordinate is \(x'=4 - x\).
Let's check using the mid - point formula. For a general point \((x,y)\) and its image \((x',y')\) after reflection over \(x = 2\), the mid - point \((\frac{x + x'}{2},\frac{y + y'}{2})\) lies on \(x = 2\), so \(\frac{x + x'}{2}=2\Rightarrow x'=4 - x\).

Answer:

1. C. \(A'(2,5),B'(-1,0),C'(-6,-2)\)
2. D. \(r_{x - axis}(x,y)=(x,-y)\)
3. Assuming \(C(-4,0)\) from the graph, \(C'(-4,4)\)
4. (Based on reflection over \(y=x\) rule)
5. B. \(x = 2\)