Question

write a rule to describe each transformation.

1. s(1,3), t(4,4), u(4,1) to s(-5,-3), t(-2,-2), u(-2,-5)
2. s(-3,-1), r(-4,3), q(0,2), p(-2,-1) to s(-2,-2), r(-3,2), q(1,1), p(-1,-2)

find the coordinates of the vertices of each figure after the given transformation.

1. translation: 2 units left q(1,-1), r(1,0), s(5,-1), t(4,-5)
2. translation: 3 units down b(-5,3), c(-3,4), d(-1,0)

Explanation:

Step1: Recall translation rule

For a translation \(a\) units left and \(b\) units down, the rule for a point \((x,y)\) is \((x - a,y - b)\).

Step2: Solve problem 10

For the transformation from \(S(1,3),T(4,4),U(4,1)\) to \(S'(- 5,-3),T'(-2,-2),U'(-2,-5)\).
For point \(S\): \(x\) - coordinate changes from \(1\) to \(-5\) (\(1-6=-5\)), \(y\) - coordinate changes from \(3\) to \(-3\) (\(3 - 6=-3\)).
The rule is \((x,y)\to(x - 6,y - 6)\).
For problem 11, from \(S(-3,-1),R(-4,3),Q(0,2),P(-2,-1)\) to \(S'(-2,-2),R'(-3,2),Q'(1,1),P'(-1,-2)\).
For point \(S\): \(x\) - coordinate changes from \(-3\) to \(-2\) (\(-3 + 1=-2\)), \(y\) - coordinate changes from \(-1\) to \(-2\) (\(-1-1=-2\)).
The rule is \((x,y)\to(x + 1,y - 1)\).

Step3: Solve problem 12

Given translation 2 units left (\(a = 2,b = 0\)), for point \(Q(1,-1)\), new coordinates are \((1-2,-1)=(-1,-1)\).
For \(R(1,0)\), new coordinates are \((1 - 2,0)=(-1,0)\).
For \(S(5,-1)\), new coordinates are \((5-2,-1)=(3,-1)\).
For \(T(4,-5)\), new coordinates are \((4 - 2,-5)=(2,-5)\).

Step4: Solve problem 13

Given translation 3 units down (\(a = 0,b = 3\)), for point \(B(-5,3)\), new coordinates are \((-5,3 - 3)=(-5,0)\).
For \(C(-3,4)\), new coordinates are \((-3,4 - 3)=(-3,1)\).
For \(D(-1,0)\), new coordinates are \((-1,0 - 3)=(-1,-3)\).

Answer:

1. Rule: \((x,y)\to(x - 6,y - 6)\)
2. Rule: \((x,y)\to(x + 1,y - 1)\)
3. \(Q'(-1,-1),R'(-1,0),S'(3,-1),T'(2,-5)\)
4. \(B'(-5,0),C'(-3,1),D'(-1,-3)\)