Question

triangle tvw is dilated according to the rule
$d_{0,\frac{3}{4}}(x,y)\to(\frac{3}{4}x,\frac{3}{4}y)$ to create the image triangle
tvw, which is not shown.
what are the coordinates of the endpoints of the
segment tv?
t(-3,6) and v(0,3)
t(-3,6) and v(0,1)
t(-1,2) and v(0,3)
t(-1,2) and v(0,1)

Explanation:

Step1: Find coordinates of T

Original coordinates of T: (-4, 8).
Using dilation rule \(D_{O,\frac{3}{4}}(x,y)\to(\frac{3}{4}x,\frac{3}{4}y)\)
\(x\) - coordinate of \(T'\): \(\frac{3}{4}\times(- 4)=-1\)
\(y\) - coordinate of \(T'\): \(\frac{3}{4}\times8 = 2\)
So \(T'(-1,2)\)

Step2: Find coordinates of V

Original coordinates of V: (0, 4).
Using dilation rule \(D_{O,\frac{3}{4}}(x,y)\to(\frac{3}{4}x,\frac{3}{4}y)\)
\(x\) - coordinate of \(V'\): \(\frac{3}{4}\times0 = 0\)
\(y\) - coordinate of \(V'\): \(\frac{3}{4}\times4=1\)
So \(V'(0,1)\)

Answer:

T'(-1, 2) and V'(0, 1)