Question

1. graph quadrilateral wxyz with vertices w(-4, 0), x( 1,2), y(-1,4), and z (-5,2) and its image after the given glide reflection: translation: using the vector < -2,3 > reflection: over the line y = - 1 w (-4,0) → w ( , ) → w ( , ) x (1,2) → x ( , ) → x ( , ) y (-1,4) → y ( , ) → y ( , ) z (-5,2) → z ( , ) → z ( , )

Explanation:

Step1: Translate point W(-4,0)

To translate a point \((x,y)\) using vector \(\langle a,b
angle\), we use the formula \((x + a, y + b)\). For \(W(-4,0)\) and vector \(\langle -2,3
angle\), we calculate \(x'=-4+(-2)= -6\) and \(y' = 0 + 3=3\). So \(W'(-6,3)\).

Step2: Reflect \(W'\) over \(y = - 1\)

The formula for reflecting a point \((x,y)\) over the line \(y = k\) is \((x,2k - y)\). Here \(k=-1\), so for \(W'(-6,3)\), \(y''=2(-1)-3=-2 - 3=-5\) and \(x''=-6\). So \(W''(-6,-5)\).

Step3: Translate point X(1,2)

Using the translation formula \((x + a, y + b)\) with \(a=-2\), \(b = 3\) and \(x = 1\), \(y = 2\). \(x'=1+(-2)=-1\), \(y'=2 + 3=5\). So \(X'(-1,5)\).

Step4: Reflect \(X'\) over \(y=-1\)

Using the reflection formula \((x,2k - y)\) with \(k = - 1\) and \((x,y)=(-1,5)\). \(y''=2(-1)-5=-2-5=-7\), \(x''=-1\). So \(X''(-1,-7)\).

Step5: Translate point Y(-1,4)

Using translation formula: \(x'=-1+(-2)=-3\), \(y'=4 + 3=7\). So \(Y'(-3,7)\).

Step6: Reflect \(Y'\) over \(y=-1\)

Using reflection formula: \(y''=2(-1)-7=-2 - 7=-9\), \(x''=-3\). So \(Y''(-3,-9)\).

Step7: Translate point Z(-5,2)

Using translation formula: \(x'=-5+(-2)=-7\), \(y'=2 + 3=5\). So \(Z'(-7,5)\).

Step8: Reflect \(Z'\) over \(y=-1\)

Using reflection formula: \(y''=2(-1)-5=-2-5=-7\), \(x''=-7\). So \(Z''(-7,-7)\).

Answer:

\(W(-4,0)\to W'(-6,3)\to W''(-6,-5)\)

\(X(1,2)\to X'(-1,5)\to X''(-1,-7)\)

\(Y(-1,4)\to Y'(-3,7)\to Y''(-3,-9)\)

\(Z(-5,2)\to Z'(-7,5)\to Z''(-7,-7)\)