To solve this problem of dilating the figure by a factor of $\frac{4}{3}$ centered at the origin, we follow these steps:

Key Concept:

1. Dilation formula for a point $(x, y)$ centered at the origin with scale factor $k$: $(x', y') = (k \cdot x, k \cdot y)$ Here, $k = \frac{4}{3}$.

2. The coordinates of each vertex of the given figure will be scaled using this formula.

Step 1: Identify coordinates of the vertices

From the graph, the coordinates appear to be:

• $H(-6, 0)$
• $I(-3, -3)$
• $J(3, -3)$
• $K(1, 3)$
• $L(0, 6)$

Step 2: Apply dilation $\frac{4}{3}$

Multiply each coordinate by $\frac{4}{3}$:

• H: $(-6, 0) \rightarrow \left( \frac{4}{3} \cdot -6, \frac{4}{3} \cdot 0 \right) = (-8, 0)$
• I: $(-3, -3) \rightarrow \left( \frac{4}{3} \cdot -3, \frac{4}{3} \cdot -3 \right) = (-4, -4)$
• J: $(3, -3) \rightarrow \left( \frac{4}{3} \cdot 3, \frac{4}{3} \cdot -3 \right) = (4, -4)$
• K: $(1, 3) \rightarrow \left( \frac{4}{3} \cdot 1, \frac{4}{3} \cdot 3 \right) = \left( \frac{4}{3}, 4 \right)$
• L: $(0, 6) \rightarrow \left( \frac{4}{3} \cdot 0, \frac{4}{3} \cdot 6 \right) = (0, 8)$

Step 3: Plot the new coordinates

The resulting points after dilation are:

• $H'(-8, 0)$
• $I'(-4, -4)$
• $J'(4, -4)$
• $K'\left( \frac{4}{3}, 4 \right)$ → approximately $(1.33, 4)$
• $L'(0, 8)$

Step 4: Draw the image

1. Plot the new points $H', I', J', K', L'$ on the coordinate plane.
2. Connect the points in the same order as the original figure.

Final Result:

The new, dilated image is a larger version of the original figure, scaled by a factor of $\frac{4}{3}$ and centered at the origin.

Do you need a graphing verification or further details?

Related Questions:

1. What happens if the dilation factor is less than 1?
2. How do you perform a dilation not centered at the origin?
3. What are the effects of a negative scale factor in dilation?
4. How would you write the general formula for dilations in vector form?
5. How can you use dilations to prove two figures are similar?

Tip: Always confirm the center of dilation before applying the scale factor. It impacts all points relative to that center.