The options of the question are the points
\(A(-5,-3)\\B(3,-3)\\C(-1,-6)\\D(4,-9)\)
Step 1
Find the rule of the translation of the pre-image to the image
we know that
the point S has coordinates of \((3,-5)\)
the point S' has coordinates of \((-4,1)\)
so
a) the rule of the translation of the pre-image to the image is
\((x.y)----> (x-7,y+6)\)
that means
the translation is \(7\) units to the left and \(6\) units up
Step 2
Find the inverse rule of the translation of the image to the pre-image
a) the inverse rule of the translation of the image to the pre-image is
\((x',y')---> (x'+7,y'-6)\)
that means
the translation is \(7\) units to the right and \(6\) units down
Step 3
Find the coordinates of the vertices of the pre-image
Applying the inverse rule of the translation of the image to the pre-image
a) Point \(R'(-8,1)\)
\((x',y')---> (x'+7,y'-6)\)
\(R'(-8,1)---> R(-8+7,1-6)\)
\(R'(-8,1)---> R(-1,-5)\)
b) Point \(T'(-4,-3)\)
\((x',y')---> (x'+7,y'-6)\)
\(T'(-4,-3)---> T(-4+7,-3-6)\)
\(T'(-4,-3)---> T(3,-9)\)
c) Point \(U'(-8,-3)\)
\((x',y')---> (x'+7,y'-6)\)
\(U'(-8,-3)---> U(-8+7,-3-6)\)
\(U'(-8,-3)---> U(-1,-9)\)
Step 4
Using a graphing tool
graph the points of the pre-image and the points A,B,C and D to determine the solution of the problem
we have
\(R(-1,-5)\\S(3,-5)\\T(3,-9)\\U(-1,-9)\)
\(A(-5,-3)\\B(3,-3)\\C(-1,-6)\\D(4,-9)\)
see the attached figure
therefore
the answer is
\(C(-1,-6)\)
