Question

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

1. translation: ((x,y)\to(x + 1,y + 5))

(n(-4,-2),v(-1,-1),q(-4,-5))

1. translation: ((x,y)\to(x + 1,y))

(d(-2,-2),n(-3,0),k(0,1),r(2,-3))

1. translation: ((x,y)\to(x - 2,y - 4))

(x(-1,1),k(3,3),t(3,-1))

1. translation: ((x,y)\to(x + 5,y + 1))

(z(-3,-4),k(-2,-1),i(0,-1))

Explanation:

Step1: For translation \((x,y)\to(x + a,y + b)\)

For each vertex \((x,y)\) of the figure, we add \(a\) to the \(x -\)coordinate and \(b\) to the \(y -\)coordinate.

Problem 13:

Given translation \((x,y)\to(x + 1,y+5)\) and vertices \(N(-4,-2)\), \(V(-1,-1)\), \(Q(-4,-5)\)

• For \(N(-4,-2)\):

\(x=-4\), \(y = - 2\)
New \(x\) - coordinate: \(x+1=-4 + 1=-3\)
New \(y\) - coordinate: \(y + 5=-2+5 = 3\)
So \(N'(-3,3)\)

• For \(V(-1,-1)\):

\(x=-1\), \(y=-1\)
New \(x\) - coordinate: \(x + 1=-1+1=0\)
New \(y\) - coordinate: \(y + 5=-1 + 5=4\)
So \(V'(0,4)\)

• For \(Q(-4,-5)\):

\(x=-4\), \(y=-5\)
New \(x\) - coordinate: \(x + 1=-4+1=-3\)
New \(y\) - coordinate: \(y + 5=-5+5 = 0\)
So \(Q'(-3,0)\)

Problem 14:

Given translation \((x,y)\to(x + 1,y)\) and vertices \(D(-2,-2)\), \(N(-3,0)\), \(K(0,1)\), \(R(2,-3)\)

• For \(D(-2,-2)\):

\(x=-2\), \(y=-2\)
New \(x\) - coordinate: \(x + 1=-2+1=-1\)
New \(y\) - coordinate: \(y=-2\)
So \(D'(-1,-2)\)

• For \(N(-3,0)\):

\(x=-3\), \(y = 0\)
New \(x\) - coordinate: \(x + 1=-3+1=-2\)
New \(y\) - coordinate: \(y = 0\)
So \(N'(-2,0)\)

• For \(K(0,1)\):

\(x=0\), \(y = 1\)
New \(x\) - coordinate: \(x + 1=0+1=1\)
New \(y\) - coordinate: \(y = 1\)
So \(K'(1,1)\)

• For \(R(2,-3)\):

\(x=2\), \(y=-3\)
New \(x\) - coordinate: \(x + 1=2+1=3\)
New \(y\) - coordinate: \(y=-3\)
So \(R'(3,-3)\)

Problem 15:

Given translation \((x,y)\to(x-2,y - 4)\) and vertices \(X(-1,1)\), \(K(3,3)\), \(T(3,-1)\)

• For \(X(-1,1)\):

\(x=-1\), \(y = 1\)
New \(x\) - coordinate: \(x-2=-1-2=-3\)
New \(y\) - coordinate: \(y-4=1 - 4=-3\)
So \(X'(-3,-3)\)

• For \(K(3,3)\):

\(x=3\), \(y = 3\)
New \(x\) - coordinate: \(x-2=3-2=1\)
New \(y\) - coordinate: \(y-4=3-4=-1\)
So \(K'(1,-1)\)

• For \(T(3,-1)\):

\(x=3\), \(y=-1\)
New \(x\) - coordinate: \(x-2=3-2=1\)
New \(y\) - coordinate: \(y-4=-1-4=-5\)
So \(T'(1,-5)\)

Problem 16:

Given translation \((x,y)\to(x + 5,y+1)\) and vertices \(Z(-3,-4)\), \(K(-2,-1)\), \(I(0,-1)\)

• For \(Z(-3,-4)\):

\(x=-3\), \(y=-4\)
New \(x\) - coordinate: \(x + 5=-3+5=2\)
New \(y\) - coordinate: \(y + 1=-4+1=-3\)
So \(Z'(2,-3)\)

• For \(K(-2,-1)\):

\(x=-2\), \(y=-1\)
New \(x\) - coordinate: \(x + 5=-2+5=3\)
New \(y\) - coordinate: \(y + 1=-1+1=0\)
So \(K'(3,0)\)

• For \(I(0,-1)\):

\(x=0\), \(y=-1\)
New \(x\) - coordinate: \(x + 5=0+5=5\)
New \(y\) - coordinate: \(y + 1=-1+1=0\)
So \(I'(5,0)\)

Answer:

1. \(N'(-3,3)\), \(V'(0,4)\), \(Q'(-3,0)\)
2. \(D'(-1,-2)\), \(N'(-2,0)\), \(K'(1,1)\), \(R'(3,-3)\)
3. \(X'(-3,-3)\), \(K'(1,-1)\), \(T'(1,-5)\)
4. \(Z'(2,-3)\), \(K'(3,0)\), \(I'(5,0)\)