Solved on Jan 18, 2024

Plot the image of a triangle with vertices at $(-3,6)$ , $(6,0)$ , and $(-6,-9)$ dilated by a factor of $\frac{1}{3}$ centered at the origin.

STEP 1

Assumptions
1. The original figure is a triangle with vertices at $(-3,6)$ , $(6,0)$ , and $(-6,-9)$ .
2. The dilation factor is $\frac{1}{3}$ .
3. The center of dilation is the origin $(0,0)$ .

STEP 2

To dilate the figure, we multiply the coordinates of each vertex by the dilation factor.
$New\, vertex = Dilation\, factor \times Original\, vertex$

STEP 3

Dilate the first vertex $(-3,6)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, A = \frac{1}{3} \times (-3,6)$

STEP 4

Calculate the coordinates of the new vertex $A$ .
$New\, vertex\, A = \left(\frac{1}{3} \times -3, \frac{1}{3} \times 6\right)$
$New\, vertex\, A = (-1,2)$

STEP 5

Dilate the second vertex $(6,0)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, B = \frac{1}{3} \times (6,0)$

STEP 6

Calculate the coordinates of the new vertex $B$ .
$New\, vertex\, B = \left(\frac{1}{3} \times 6, \frac{1}{3} \times 0\right)$
$New\, vertex\, B = (2,0)$

STEP 7

Dilate the third vertex $(-6,-9)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, C = \frac{1}{3} \times (-6,-9)$

STEP 8

Calculate the coordinates of the new vertex $C$ .
$New\, vertex\, C = \left(\frac{1}{3} \times -6, \frac{1}{3} \times -9\right)$
$New\, vertex\, C = (-2,-3)$

STEP 9

Now that we have the new vertices, we can plot the resulting image. The new triangle will have vertices at $(-1,2)$ , $(2,0)$ , and $(-2,-3)$ .
The dilation process has been completed, and the resulting triangle is smaller by a factor of $\frac{1}{3}$ compared to the original triangle. To visualize the result, you can draw the original triangle on a coordinate plane and then plot the new vertices, connecting them to form the dilated triangle.

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Plot the image of a triangle with vertices at $(-3,6)$ , $(6,0)$ , and $(-6,-9)$ dilated by a factor of $\frac{1}{3}$ centered at the origin.

STEP 1

Assumptions
1. The original figure is a triangle with vertices at $(-3,6)$ , $(6,0)$ , and $(-6,-9)$ .
2. The dilation factor is $\frac{1}{3}$ .
3. The center of dilation is the origin $(0,0)$ .

STEP 2

To dilate the figure, we multiply the coordinates of each vertex by the dilation factor.
$New\, vertex = Dilation\, factor \times Original\, vertex$

STEP 3

Dilate the first vertex $(-3,6)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, A = \frac{1}{3} \times (-3,6)$

STEP 4

Calculate the coordinates of the new vertex $A$ .
$New\, vertex\, A = \left(\frac{1}{3} \times -3, \frac{1}{3} \times 6\right)$
$New\, vertex\, A = (-1,2)$

STEP 5

Dilate the second vertex $(6,0)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, B = \frac{1}{3} \times (6,0)$

STEP 6

Calculate the coordinates of the new vertex $B$ .
$New\, vertex\, B = \left(\frac{1}{3} \times 6, \frac{1}{3} \times 0\right)$
$New\, vertex\, B = (2,0)$

STEP 7

Dilate the third vertex $(-6,-9)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, C = \frac{1}{3} \times (-6,-9)$

STEP 8

Calculate the coordinates of the new vertex $C$ .
$New\, vertex\, C = \left(\frac{1}{3} \times -6, \frac{1}{3} \times -9\right)$
$New\, vertex\, C = (-2,-3)$

STEP 9

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Plot the image of a triangle with vertices at (−3,6)(-3,6)(−3,6), (6,0)(6,0)(6,0), and (−6,−9)(-6,-9)(−6,−9) dilated by a factor of 13\frac{1}{3}31​ centered at the origin.

STEP 1

Assumptions1. The original figure is a triangle with vertices at (−3,6)(-3,6)(−3,6), (6,0)(6,0)(6,0), and (−6,−9)(-6,-9)(−6,−9).2. The dilation factor is 13\frac{1}{3}31​.3. The center of dilation is the origin (0,0)(0,0)(0,0).

STEP 2

To dilate the figure, we multiply the coordinates of each vertex by the dilation factor.New vertex=Dilation factor×Original vertexNew\, vertex = Dilation\, factor \times Original\, vertexNewvertex=Dilationfactor×Originalvertex

STEP 3

Dilate the first vertex (−3,6)(-3,6)(−3,6) by the dilation factor 13\frac{1}{3}31​.New vertex A=13×(−3,6)New\, vertex\, A = \frac{1}{3} \times (-3,6)NewvertexA=31​×(−3,6)

STEP 4

Calculate the coordinates of the new vertex AAA.New vertex A=(13×−3,13×6)New\, vertex\, A = \left(\frac{1}{3} \times -3, \frac{1}{3} \times 6\right)NewvertexA=(31​×−3,31​×6)New vertex A=(−1,2)New\, vertex\, A = (-1,2)NewvertexA=(−1,2)

STEP 5

Dilate the second vertex (6,0)(6,0)(6,0) by the dilation factor 13\frac{1}{3}31​.New vertex B=13×(6,0)New\, vertex\, B = \frac{1}{3} \times (6,0)NewvertexB=31​×(6,0)

STEP 6

Calculate the coordinates of the new vertex BBB.New vertex B=(13×6,13×0)New\, vertex\, B = \left(\frac{1}{3} \times 6, \frac{1}{3} \times 0\right)NewvertexB=(31​×6,31​×0)New vertex B=(2,0)New\, vertex\, B = (2,0)NewvertexB=(2,0)

STEP 7

Dilate the third vertex (−6,−9)(-6,-9)(−6,−9) by the dilation factor 13\frac{1}{3}31​.New vertex C=13×(−6,−9)New\, vertex\, C = \frac{1}{3} \times (-6,-9)NewvertexC=31​×(−6,−9)

STEP 8

Calculate the coordinates of the new vertex CCC.New vertex C=(13×−6,13×−9)New\, vertex\, C = \left(\frac{1}{3} \times -6, \frac{1}{3} \times -9\right)NewvertexC=(31​×−6,31​×−9)New vertex C=(−2,−3)New\, vertex\, C = (-2,-3)NewvertexC=(−2,−3)

STEP 9

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Plot the image of a triangle with vertices at $(-3,6)$ , $(6,0)$ , and $(-6,-9)$ dilated by a factor of $\frac{1}{3}$ centered at the origin.

Assumptions
1. The original figure is a triangle with vertices at $(-3,6)$ , $(6,0)$ , and $(-6,-9)$ .
2. The dilation factor is $\frac{1}{3}$ .
3. The center of dilation is the origin $(0,0)$ .

To dilate the figure, we multiply the coordinates of each vertex by the dilation factor.
$New\, vertex = Dilation\, factor \times Original\, vertex$

Dilate the first vertex $(-3,6)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, A = \frac{1}{3} \times (-3,6)$

Calculate the coordinates of the new vertex $A$ .
$New\, vertex\, A = \left(\frac{1}{3} \times -3, \frac{1}{3} \times 6\right)$
$New\, vertex\, A = (-1,2)$

Dilate the second vertex $(6,0)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, B = \frac{1}{3} \times (6,0)$

Calculate the coordinates of the new vertex $B$ .
$New\, vertex\, B = \left(\frac{1}{3} \times 6, \frac{1}{3} \times 0\right)$
$New\, vertex\, B = (2,0)$

Dilate the third vertex $(-6,-9)$ by the dilation factor $\frac{1}{3}$ .
$New\, vertex\, C = \frac{1}{3} \times (-6,-9)$

Calculate the coordinates of the new vertex $C$ .
$New\, vertex\, C = \left(\frac{1}{3} \times -6, \frac{1}{3} \times -9\right)$
$New\, vertex\, C = (-2,-3)$