Question

quadrilateral abcd has vertices a(-3,1), b(-5,3), c(-2,5), and d(-1,3). what are the coordinates of the vertices of the image a translation along vector (5, -3)? a. a(2, -2), b(0, 0), c(3, 2), and d(4, 0) b. a(-8, -2), b(-10, 0), c(-7, 2), and d(-6, 0) c. a(2, 4), b(0, 6), c(3, 8), and d(4, 6) d. a(-8, 4), b(-10, 6), c(-7, 8), and d(-6, 6)

Explanation:

Step1: Apply translation rule

For a point \((x,y)\) translated along vector \((a,b)\), the new coordinates are \((x + a,y + b)\). Here \(a = 5\) and \(b=-3\).
For point \(A(-3,1)\):
\(x=-3,y = 1\)
\(x'=-3 + 5=2\)
\(y'=1+( - 3)=-2\)
So \(A'(2,-2)\)

Step2: For point \(B(-5,3)\)

\(x=-5,y = 3\)
\(x'=-5 + 5=0\)
\(y'=3+( - 3)=0\)
So \(B'(0,0)\)

Step3: For point \(C(-2,5)\)

\(x=-2,y = 5\)
\(x'=-2 + 5=3\)
\(y'=5+( - 3)=2\)
So \(C'(3,2)\)

Step4: For point \(D(-1,3)\)

\(x=-1,y = 3\)
\(x'=-1 + 5=4\)
\(y'=3+( - 3)=0\)
So \(D'(4,0)\)

Answer:

A. \(A'(2, - 2),B'(0,0),C'(3,2)\), and \(D'(4,0)\)