Respuesta :
To dilate of the small triangle to get the large triangle the center of dilation is required
The description of the dilation is as follows;
Location of the center of dilation;
The coordinate of the center of dilation is given as follows;
[tex]\left( x_0 = \dfrac{k \cdot x_1 - x_2}{k - 1}, \ y_0 = \dfrac{k \cdot y_1 - y_2}{k - 1} \right)[/tex]
Where;
k = The scale factor of dilation
(x₁, y₁) = A coordinate point on the preimage
(x₂, y₂) = A coordinate point on the image
From the given drawing, we have;
Two vertices on the preimage are; (2, 0), and (5, 1)
Two corresponding vertices on the image are; (0, 0), and (9, 3)
Using the formula for length, l, we have;
[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex]
[tex]l_{small} = \sqrt{\left (1 -0 \right )^{2}+\left (5-2 \right )^{2}} = \sqrt{10}[/tex]
[tex]l_{large} = \sqrt{\left (9 -0 \right )^{2}+\left (3-0 \right )^{2}} = \sqrt{90} = 3 \cdot \sqrt{10}[/tex]
- [tex]The \ scale \ factor, \ k = \dfrac{l_{large}}{l_{small}} = \dfrac{3 \cdot \sqrt{10} }{\sqrt{10} } = 3[/tex]
[tex]\left( x_0 = \dfrac{3 \times 2 -0}{3 - 1}, \ y_0 = \dfrac{3 \times 0 - 0}{3 - 1} \right)[/tex]
[tex]\left( x_0 = 3, \ y_0 = 0\right)[/tex]
- The center of dilation is the point (3, 0)
The steps to create the larger triangle from the center of dilation are as follows;
- From the center of dilation, draw lines passing through the vertices of the small triangle
- With the ruler mark points on the radial lines that are three times the distances from the vertices of the small triangle from the center of dilation
- Join the marked points to create the larger triangle
Alternatively larger triangle can also be created using the vertices as follows;
The coordinates of the third vertex of the small triangle are; (4, -2)
To obtain the coordinates of the large triangle, the coordinates of
the small triangle are multiplied by the scale factor, followed by adding the
coordinates of the center of dilation, we have that the coordinates of the
image are;
[tex](k \times x_{ai} + x_0, \ k \times y_{ai} + y_0)[/tex]
Where;
[tex](x_{ai}, \ y_{ai})[/tex] = The difference between the coordinate values of the vertices of
the small triangle and the center of dilation, we have;
(2 - 3, 0 - 0) = (-1, 0)
(5 - 3, 1 - 0) = (2, 1)
(4 - 3, -2 - 0) = (1, -2)
We have [tex](k \times x_{ai} + x_0, \ k \times y_{ai} + y_0)[/tex] are;
(3 ×(-1) + 3, 3× 0 + 3) = (-3 + 3, 0) = (0, 0)
(3 × 2 + 3, 3 × 1 + 0) = (6 + 3, 3) = (9, 3)
(3 × 1, 3 × (-2) + 0) = (3 + 3, -6) = (6, -6)
The vertices of the larger triangle are found as (0, 0), (9, 3), (6, -6)
By drawing lines joining derived vertices the larger triangle can be created
Learn more about scale drawing here:
https://brainly.com/question/20418185
