Question

original coordinates → final coordinates
$a(-4,6)→a(-6,9)$
$b(6,2)→b(9,3)$
$c(-6,-4)→c(-9,-6)$
(b) choose the general rule below that describes the dilation mapping $△abc$ to $△abc$.
$\bigcirc (x,y)→(\frac {2}{3}x,\frac {3}{2}y)$ $\bigcirc (x,y)→(\frac {2}{3}x,\frac {2}{3}y)$
$\bigcirc (x,y)→(\frac {3}{2}y,\frac {3}{2}x)$ $\bigcirc (x,y)→(\frac {3}{2}x,y)$
$\bigcirc (x,y)→(\frac {3}{2}x,\frac {3}{2}y)$ $\bigcirc (x,y)→(\frac {3}{2}x,\frac {2}{3}y)$
$\bigcirc (x,y)→(x,\frac {3}{2}y)$ $\bigcirc (x,y)→(\frac {2}{3}y,\frac {2}{3}x)$

Explanation:

Step1: Analyze the transformation of \(x\) - coordinate

Take point \(A(-4,6)\) and \(A'(-6,9)\). For the \(x\) - coordinate: \(\frac{-6}{-4}=\frac{3}{2}\). Take point \(B(6,2)\) and \(B'(9,3)\), \(\frac{9}{6}=\frac{3}{2}\). Take point \(C(-6,-4)\) and \(C'(-9,-6)\), \(\frac{-9}{-6}=\frac{3}{2}\).

Step2: Analyze the transformation of \(y\) - coordinate

For point \(A(-4,6)\) and \(A'(-6,9)\), \(\frac{9}{6}=\frac{3}{2}\). For point \(B(6,2)\) and \(B'(9,3)\), \(\frac{3}{2}=\frac{3}{2}\). For point \(C(-6,-4)\) and \(C'(-9,-6)\), \(\frac{-6}{-4}=\frac{3}{2}\).

Answer:

\((x,y)\to(\frac{3}{2}x,\frac{3}{2}y)\)