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? o t(-3,6) and v(0,2) o t(-3,6) and v(0,1) o t(-1,2) and v(0,3) o t(-1,2) and v(0,1)

Explanation:

Step1: Assume original - point coordinates

Let's assume we reverse - dilate the options using the inverse of the dilation rule $D_{0,\frac{3}{4}}(x,y)=(\frac{3}{4}x,\frac{3}{4}y)$. The inverse rule is $D^{-1}(x,y)=(\frac{4}{3}x,\frac{4}{3}y)$.

Step2: Check each option

For option B: If $T(-3,6)$ and $V(0,1)$, after applying $D_{0,\frac{3}{4}}$, for $T$: $(\frac{3}{4}\times(-3),\frac{3}{4}\times6)=(- \frac{9}{4},\frac{9}{2})$. For $V$: $(\frac{3}{4}\times0,\frac{3}{4}\times1)=(0,\frac{3}{4})$.
For option D: If $T(-1,2)$ and $V(0,1)$, for $T$: $(\frac{3}{4}\times(-1),\frac{3}{4}\times2)=(-\frac{3}{4},\frac{3}{2})$. For $V$: $(\frac{3}{4}\times0,\frac{3}{4}\times1)=(0,\frac{3}{4})$.
For option C: If $T(-1,2)$ and $V(0,3)$, for $T$: $(\frac{3}{4}\times(-1),\frac{3}{4}\times2)=(-\frac{3}{4},\frac{3}{2})$. For $V$: $(\frac{3}{4}\times0,\frac{3}{4}\times3)=(0,\frac{9}{4})$.
For option A: If $T(-3,6)$ and $V(0,2)$, for $T$: $(\frac{3}{4}\times(-3),\frac{3}{4}\times6)=(-\frac{9}{4},\frac{9}{2})$. For $V$: $(\frac{3}{4}\times0,\frac{3}{4}\times2)=(0,\frac{3}{2})$.
If we assume the original triangle's coordinates and then apply the dilation rule $D_{0,\frac{3}{4}}$, we find that if the original endpoints of $TV$ are $T(- 4,8)$ and $V(0,\frac{8}{3})$, after dilation $D_{0,\frac{3}{4}}$, for $T$: $(\frac{3}{4}\times(-4),\frac{3}{4}\times8)=(-3,6)$ and for $V$: $(\frac{3}{4}\times0,\frac{3}{4}\times\frac{8}{3})=(0,2)$.

Answer:

A. T(-3,6) and V(0,2)