Question

1. the coordinates of ( \triangle abc ) are ( a(2,3) ), ( b(6,6) ), and ( c(7,2) ). after a translation, the image of vertex ( a ) is ( a(-6,1) ).

a) give the coordinates of ( b ) and ( c ) after the same translation.

b) graph ( \triangle abc ) and ( \triangle abc ) on the graph above and to the right.

1. the coordinates of ( defg ) are ( d(2,-1) ), ( e(5,-2) ), ( f(4,-5) ), and ( g(1,-4) ).

a) describe in coordinate mapping notation a translation that will move vertex ( e ) to the origin.

b) give the coordinates of ( d ), ( e ), ( f ), and ( g ) after the translation described in part a).

(use your own graph paper for b, c, and

Explanation:

Step1: Find the translation rule for problem 6a

To find the translation rule from \(A(2,3)\) to \(A'(-6,1)\), we use the formula \((x,y)\to(x + a,y + b)\).
For the \(x\)-coordinate: \(2+a=-6\), so \(a=-6 - 2=-8\).
For the \(y\)-coordinate: \(3 + b=1\), so \(b=1 - 3=-2\).
The translation rule is \((x,y)\to(x-8,y - 2)\).
For \(B(6,6)\):
\(x\)-coordinate of \(B'\): \(6-8=-2\)
\(y\)-coordinate of \(B'\): \(6-2 = 4\)
For \(C(7,2)\):
\(x\)-coordinate of \(C'\): \(7-8=-1\)
\(y\)-coordinate of \(C'\): \(2-2=0\)

Step2: Find the translation rule for problem 7a

We want to move \(E(5,-2)\) to \((0,0)\).
Using the formula \((x,y)\to(x + a,y + b)\), for \(x\)-coordinate: \(5+a = 0\Rightarrow a=-5\), for \(y\)-coordinate: \(-2 + b=0\Rightarrow b = 2\).
The translation rule is \((x,y)\to(x-5,y + 2)\)

Step3: Find the coordinates for problem 7b

For \(D(2,-1)\):
\(x\)-coordinate of \(D'\): \(2-5=-3\)
\(y\)-coordinate of \(D'\): \(-1 + 2=1\)
For \(E(5,-2)\): \((0,0)\) (already known from the rule)
For \(F(4,-5)\):
\(x\)-coordinate of \(F'\): \(4-5=-1\)
\(y\)-coordinate of \(F'\): \(-5 + 2=-3\)
For \(G(1,-4)\):
\(x\)-coordinate of \(G'\): \(1-5=-4\)
\(y\)-coordinate of \(G'\): \(-4+2=-2\)

Answer:

6a. \(B'(-2,4)\), \(C'(-1,0)\)
7a. \((x,y)\to(x - 5,y + 2)\)
7b. \(D'(-3,1)\), \(E'(0,0)\), \(F'(-1,-3)\), \(G'(-4,-2)\)