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 \(T\) and \(V\)

From the graph, \(T(-4,8)\) and \(V(0,4)\)

Step2: Apply dilation rule \(D_{O,\frac{3}{4}}(x,y)\to(\frac{3}{4}x,\frac{3}{4}y)\)

For \(T(-4,8)\):
\(x = - 4,y = 8\)
\(x'=\frac{3}{4}\times(-4)=-3\), \(y'=\frac{3}{4}\times8 = 6\)
So \(T'(-3,6)\)
For \(V(0,4)\):
\(x = 0,y = 4\)
\(x'=\frac{3}{4}\times0=0\), \(y'=\frac{3}{4}\times4 = 3\)
So \(V'(0,3)\)

Answer:

T'(-3, 6) and V'(0, 3)