Question

graph the image of $\triangle tuv$ after a dilation with a scale factor of $\frac{1}{2}$, centered at the origin.

Explanation:

Step1: Identify coordinates of ΔTUV

From the graph: \( T(0,6) \), \( U(8,6) \), \( V(-6,-8) \)

Step2: Apply dilation (scale factor \( \frac{1}{2} \), origin center)

For each point \((x,y)\), new point is \((\frac{1}{2}x, \frac{1}{2}y)\):
\( T'(0 \times \frac{1}{2}, 6 \times \frac{1}{2}) = (0,3) \)
\( U'(8 \times \frac{1}{2}, 6 \times \frac{1}{2}) = (4,3) \)
\( V'(-6 \times \frac{1}{2}, -8 \times \frac{1}{2}) = (-3,-4) \)

Step3: Plot new triangle ΔT'U'V'

Connect \( T'(0,3) \), \( U'(4,3) \), \( V'(-3,-4) \) on the grid.

Answer:

The image of \( \Delta TUV \) after dilation has vertices \( T'(0,3) \), \( U'(4,3) \), \( V'(-3,-4) \); plot these points and connect them to form the dilated triangle.