Question

complete the table to show the coordinates of the vertices of △abc. △abc a(-2,3) b(0,4) c(1,1) △abc a(1,3) ? ? c(1,3) c(4,1) c(3,4)

Explanation:

Step1: Find the transformation rule

Compare the coordinates of \(A(-2,3)\) and \(A'(1,3)\). The \(y -\)coordinate remains the same (\(y = 3\)), and the \(x -\)coordinate changes from \(-2\) to \(1\). The transformation for the \(x -\)coordinate is \(x'=x + 3\) (since \(-2+3 = 1\)).

Step2: Apply the transformation rule to point \(B\)

For point \(B(0,4)\), using the rule \(x'=x + 3\) and \(y'=y\).
The \(x -\)coordinate of \(B'\) is \(x=0+3=3\), and the \(y -\)coordinate of \(B'\) is \(y = 4\). So \(B'(3,4)\).

Step3: Apply the transformation rule to point \(C\)

For point \(C(1,1)\), using the rule \(x'=x + 3\) and \(y'=y\).
The \(x -\)coordinate of \(C'\) is \(x=1 + 3=4\), and the \(y -\)coordinate of \(C'\) is \(y = 1\). But wait, no, we made a mistake. Wait, looking at \(A(-2,3)\to A'(1,3)\) (translation \(3\) units to the right).
For \(B(0,4)\): \(x=0 + 3=3,y = 4\), so \(B'(3,4)\)
For \(C(1,1)\): \(x=1+3=4,y = 1\), but wait, no, looking at the second - row \(C'(4,1)\) is already there. Wait, no, the problem is to complete the table.
Wait, actually, from \(A(-2,3)\) to \(A'(1,3)\) is \(T:(x,y)\to(x + 3,y)\)
For \(B(0,4)\): \(B'(0 + 3,4)=(3,4)\)
For \(C(1,1)\): \(C'(1+3,1)=(4,1)\)

Answer:

\(B'(3,4)\)