Question

the vertices of △abc are a(1, 2), b(5, 1), c(5, 4). translate and its image 1) x, y)->(x - 5, y)

Explanation:

Step1: Translate point A

Given \(A(1,2)\) and the translation rule \((x,y)\to(x - 5,y)\). Substitute \(x = 1\) and \(y=2\) into the rule.
\(x'=1 - 5=-4\), \(y' = 2\). So the new - point \(A'\) is \((-4,2)\).

Step2: Translate point B

Given \(B(5,1)\) and the translation rule \((x,y)\to(x - 5,y)\). Substitute \(x = 5\) and \(y = 1\) into the rule.
\(x'=5 - 5 = 0\), \(y'=1\). So the new - point \(B'\) is \((0,1)\).

Step3: Translate point C

Given \(C(5,4)\) and the translation rule \((x,y)\to(x - 5,y)\). Substitute \(x = 5\) and \(y = 4\) into the rule.
\(x'=5 - 5=0\), \(y' = 4\). So the new - point \(C'\) is \((0,4)\).

Answer:

The vertices of the translated triangle \(\triangle A'B'C'\) are \(A'(-4,2)\), \(B'(0,1)\), \(C'(0,4)\)