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An art student uses dilations in all her art. She first plans the art piece on a coordinate grid. Determine the vertices of the image of the triangle with vertices A(1,1) B(2,4) and C(3,9) after a dilation with scale factor 1.5.

I'm very confused with this problem, so I need all the help I can get. Thanks to anybody that has helped me solve this problem! I really appreciate it :)

Guest Feb 16, 2016
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An art student uses dilations in all her art. She first plans the art piece on a coordinate grid. Determine the vertices of the image of the triangle with vertices A(1,1) B(2,4) and C(3,9) after a dilation with scale factor 1.5.

I asume the center of enlargement is the centroid.

We have: $$\begin{array}{rcll} \vec{A}=\binom{1}{1} \qquad \vec{B} = \binom{2}{4} \qquad \vec{C} = \binom{3}{9} \\ \end{array}$$

The centroid of the triangle is: $$\vec{c} = \frac13\cdot ( \vec{A}+\vec{B}+\vec{C} ) =\dbinom{\frac{x_a+x_b+x_c}{3}}{ \frac{y_a+y_b+y_c}{3}}= \dbinom{\frac{1+2+3}{3}}{ \frac{1+4+9}{3}} = \dbinom{2}{\frac{14}{3}}$$

1. Barycentric coordinates:

$$\begin{array}{rcll} \vec{A}-\vec{c} &=& \dbinom{1-2}{1-\frac{14}{3}} = \dbinom{-1}{-\frac{11}{3}}\\\\ \vec{B}-\vec{c} &=& \dbinom{2-2}{4-\frac{14}{3}} = \dbinom{0}{-\frac23}\\\\ \vec{C}-\vec{c} &=& \dbinom{3-2}{9-\frac{14}{3}} = \dbinom{1}{\frac{13}{3}} \end{array}$$

2. Scale factor 1.5

$$\begin{array}{rcll} \dbinom{-1}{-\frac{11}{3}} \cdot 1.5 &=& \dbinom{-1.5}{-\frac{11}{2}}\\\\ \dbinom{0}{-\frac23}\cdot 1.5 &=& \dbinom{0}{-1}\\\\ \dbinom{1}{\frac{13}{3}}\cdot 1.5 &=& \dbinom{1.5}{\frac{13}{2}} \end{array}$$

3. The vertices of the image $$+\vec{c}$$

$$\begin{array}{rcll} \dbinom{-1.5}{-\frac{11}{2}} + \dbinom{2}{\frac{14}{3}} &=& \dbinom{0.5}{-\frac{5}{6}}\\\\ \dbinom{0}{-1} + \dbinom{2}{\frac{14}{3}} &=& \dbinom{2}{\frac{11}{3}}\\\\ \dbinom{1.5}{\frac{13}{2}} + \dbinom{2}{\frac{14}{3}} &=& \dbinom{3.5}{\frac{67}{6}} \end{array}$$

The vertices of the image are $$\begin{array}{rcll} A' (0.5,-\frac{5}{6} ) \qquad B' ( 2, \frac{11}{3}) \qquad C' ( 3.5, \frac{67}{6} ) \end{array}$$

heureka  Feb 16, 2016

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