Question

1. dilation of $\frac{3}{2}$ about the origin

Explanation:

1. First, assume the coordinates of the vertices of the original - triangle:

• Let's assume the coordinates of the vertices of \(\triangle DEF\) are \(D(x_1,y_1)\), \(E(x_2,y_2)\), and \(F(x_3,y_3)\). From the grid, if we assume \(D(0,0)\), \(E(0,4)\), and \(F(2,0)\).
• The rule for a dilation about the origin with a scale - factor \(k\) is \((x,y)\to(kx,ky)\). Here, \(k = \frac{3}{2}\).

1. Then, find the new coordinates of the vertices after dilation:

• For point \(D(0,0)\):
• Using the dilation rule \((x,y)\to(kx,ky)\) with \(k=\frac{3}{2}\), we have \(D'( \frac{3}{2}\times0,\frac{3}{2}\times0)=(0,0)\).
• For point \(E(0,4)\):
• Substitute \(x = 0\) and \(y = 4\) into the dilation rule \((x,y)\to(kx,ky)\). We get \(E'(\frac{3}{2}\times0,\frac{3}{2}\times4)=(0,6)\).
• For point \(F(2,0)\):
• Substitute \(x = 2\) and \(y = 0\) into the dilation rule \((x,y)\to(kx,ky)\). We get \(F'(\frac{3}{2}\times2,\frac{3}{2}\times0)=(3,0)\).

Answer:

The new coordinates of the vertices of the dilated triangle are \(D'(0,0)\), \(E'(0,6)\), and \(F'(3,0)\) (assuming the original vertices \(D(0,0)\), \(E(0,4)\), \(F(2,0)\)).