Question

draw the image of $\triangle abc$ under a dilation whose center is $p$ and scale factor is $\dfrac{1}{2}$.

Explanation:

Step1: Connect P to each vertex

Draw segments $PA$, $PB$, and $PC$.

Step2: Find midpoints of segments

For each segment, locate the point that is $\frac{1}{2}$ the distance from $P$ to the vertex (this is the midpoint, since scale factor is $\frac{1}{2}$):

• Midpoint of $PA$: $A'$, where $PA' = \frac{1}{2}PA$
• Midpoint of $PB$: $B'$, where $PB' = \frac{1}{2}PB$
• Midpoint of $PC$: $C'$, where $PC' = \frac{1}{2}PC$

Step3: Connect new vertices

Draw $\triangle A'B'C'$ by connecting the three midpoints.

Answer:

The image $\triangle A'B'C'$ is a smaller triangle, with each vertex lying halfway between point $P$ and the corresponding vertex of $\triangle ABC$, and side lengths half of $\triangle ABC$: $A'B' = \frac{1}{2}AB = 2.9$, $B'C' = \frac{1}{2}BC = 6.1$, $A'C' = \frac{1}{2}AC = 2.9$. The triangle is positioned closer to center $P$ than $\triangle ABC$.