Question

draw a dilation of the figure using the given scale factor.

1. (k = 2)
2. (k=\frac{1}{4})
3. (k=\frac{1}{2})
4. (k = 1\frac{1}{2})
5. draw a shape with all straight lines (be creative) and then dilate it. graph the figure and its image on the coordinate plane below. be sure to include what your scale factor was:

Explanation:

Step1: Recall dilation rule

For a point $(x,y)$ and a scale - factor $k$, the dilated point $(x',y')$ is given by $(x',y')=(k\cdot x,k\cdot y)$ when the center of dilation is the origin.

Step2: For problem 1 ($k = 2$)

Let's assume the vertices of the triangle in the first figure are $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$. The new vertices $A'(x_1',y_1')$, $B'(x_2',y_2')$ and $C'(x_3',y_3')$ will be $A'(2x_1,2y_1)$, $B'(2x_2,2y_2)$ and $C'(2x_3,2y_3)$. Plot these new points to get the dilated triangle.

Step3: For problem 2 ($k=\frac{1}{4}$)

If the vertices of the triangle are $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$, the new vertices are $A'(\frac{1}{4}x_1,\frac{1}{4}y_1)$, $B'(\frac{1}{4}x_2,\frac{1}{4}y_2)$ and $C'(\frac{1}{4}x_3,\frac{1}{4}y_3)$. Plot these to get the dilated triangle.

Step4: For problem 3 ($k = \frac{1}{2}$)

For the third - figure with vertices of the shape say $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ and $D(x_4,y_4)$, the new vertices are $A'(\frac{1}{2}x_1,\frac{1}{2}y_1)$, $B'(\frac{1}{2}x_2,\frac{1}{2}y_2)$, $C'(\frac{1}{2}x_3,\frac{1}{2}y_3)$ and $D'(\frac{1}{2}x_4,\frac{1}{2}y_4)$. Plot them to get the dilated shape.

Step5: For problem 4 ($k = 1\frac{1}{2}=\frac{3}{2}$)

If the vertices of the shape are $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ and $D(x_4,y_4)$, the new vertices are $A'(\frac{3}{2}x_1,\frac{3}{2}y_1)$, $B'(\frac{3}{2}x_2,\frac{3}{2}y_2)$, $C'(\frac{3}{2}x_3,\frac{3}{2}y_3)$ and $D'(\frac{3}{2}x_4,\frac{3}{2}y_4)$. Plot them to get the dilated shape.

Step6: For problem 5

1. Draw a polygon (for example, a square with vertices $(1,1)$, $(1,3)$, $(3,3)$ and $(3,1)$).
2. Choose a scale - factor, say $k = 3$.
3. Calculate the new vertices using the rule $(x',y')=(k\cdot x,k\cdot y)$. For the square with vertices $(1,1)$, $(1,3)$, $(3,3)$ and $(3,1)$, the new vertices are $(3,3)$, $(3,9)$, $(9,9)$ and $(9,3)$.
4. Plot the original square and the dilated square on the coordinate plane.

Since this is a drawing problem, the answer is the set of new coordinates for each problem which are then plotted on the given coordinate - planes to get the dilated figures. For a full answer, you need to identify the original coordinates of the vertices of each figure in the problems and then apply the dilation formula as shown above to get the new coordinates for plotting.

Answer:

Step1: Recall dilation rule

For a point $(x,y)$ and a scale - factor $k$, the dilated point $(x',y')$ is given by $(x',y')=(k\cdot x,k\cdot y)$ when the center of dilation is the origin.

Step2: For problem 1 ($k = 2$)

Let's assume the vertices of the triangle in the first figure are $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$. The new vertices $A'(x_1',y_1')$, $B'(x_2',y_2')$ and $C'(x_3',y_3')$ will be $A'(2x_1,2y_1)$, $B'(2x_2,2y_2)$ and $C'(2x_3,2y_3)$. Plot these new points to get the dilated triangle.

Step3: For problem 2 ($k=\frac{1}{4}$)

If the vertices of the triangle are $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$, the new vertices are $A'(\frac{1}{4}x_1,\frac{1}{4}y_1)$, $B'(\frac{1}{4}x_2,\frac{1}{4}y_2)$ and $C'(\frac{1}{4}x_3,\frac{1}{4}y_3)$. Plot these to get the dilated triangle.

Step4: For problem 3 ($k = \frac{1}{2}$)

For the third - figure with vertices of the shape say $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ and $D(x_4,y_4)$, the new vertices are $A'(\frac{1}{2}x_1,\frac{1}{2}y_1)$, $B'(\frac{1}{2}x_2,\frac{1}{2}y_2)$, $C'(\frac{1}{2}x_3,\frac{1}{2}y_3)$ and $D'(\frac{1}{2}x_4,\frac{1}{2}y_4)$. Plot them to get the dilated shape.

Step5: For problem 4 ($k = 1\frac{1}{2}=\frac{3}{2}$)

If the vertices of the shape are $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ and $D(x_4,y_4)$, the new vertices are $A'(\frac{3}{2}x_1,\frac{3}{2}y_1)$, $B'(\frac{3}{2}x_2,\frac{3}{2}y_2)$, $C'(\frac{3}{2}x_3,\frac{3}{2}y_3)$ and $D'(\frac{3}{2}x_4,\frac{3}{2}y_4)$. Plot them to get the dilated shape.

Step6: For problem 5

1. Draw a polygon (for example, a square with vertices $(1,1)$, $(1,3)$, $(3,3)$ and $(3,1)$).
2. Choose a scale - factor, say $k = 3$.
3. Calculate the new vertices using the rule $(x',y')=(k\cdot x,k\cdot y)$. For the square with vertices $(1,1)$, $(1,3)$, $(3,3)$ and $(3,1)$, the new vertices are $(3,3)$, $(3,9)$, $(9,9)$ and $(9,3)$.
4. Plot the original square and the dilated square on the coordinate plane.

Since this is a drawing problem, the answer is the set of new coordinates for each problem which are then plotted on the given coordinate - planes to get the dilated figures. For a full answer, you need to identify the original coordinates of the vertices of each figure in the problems and then apply the dilation formula as shown above to get the new coordinates for plotting.