Question

part a:
quadrilateral abcd has coordinates a(0, 2), b(3, 4), c(4, 2), and d(3, 0).
determine the coordinates of the vertices of abcd after a translation along the vector < - 2, 3 >
a(5, 2), b(1, 7), c(2, 5), d(1, 3)
a(-2, 5), b(1, 7), c(2, 5), d(1, 3)
a(2, 5), b(1, 7), c(2, 5), d(1, 3)
a(5, 2), b(1, 7), c(2, 5), d(1, 3)

Explanation:

Step1: Translation rule

When translating a point \((x,y)\) along the vector \(\langle a,b
angle\), the new coordinates \((x',y')\) are given by \(x'=x + a\) and \(y'=y + b\). Here \(a=-2\) and \(b = 3\).

Step2: Translate point \(A(0,2)\)

For \(A(0,2)\), \(x = 0\), \(y=2\). Then \(x'=0+(-2)=-2\) and \(y'=2 + 3=5\). So \(A'(-2,5)\).

Step3: Translate point \(B(3,4)\)

For \(B(3,4)\), \(x = 3\), \(y = 4\). Then \(x'=3+(-2)=1\) and \(y'=4 + 3=7\). So \(B'(1,7)\).

Step4: Translate point \(C(4,2)\)

For \(C(4,2)\), \(x = 4\), \(y = 2\). Then \(x'=4+(-2)=2\) and \(y'=2+3 = 5\). So \(C'(2,5)\).

Step5: Translate point \(D(3,0)\)

For \(D(3,0)\), \(x = 3\), \(y = 0\). Then \(x'=3+(-2)=1\) and \(y'=0 + 3=3\). So \(D'(1,3)\).

Answer:

\(A'(-2,5)\), \(B'(1,7)\), \(C'(2,5)\), \(D'(1,3)\) (the second option in the given choices)