+0

0
474
1

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
Sort:

#1
+19207
+5

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

### 11 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details