Question

write the coordinates of the vertices after a dilation with a scale factor of $\frac{1}{3}$, centered at the origin.

Explanation:

Step1: Identify original coordinates

Let's assume the vertices are \(S(0, - 3)\), \(T(0,9)\), \(U(3,9)\), \(V(3,-9)\). The formula for dilation centered at the origin with scale - factor \(k\) is \((x,y)\to(kx,ky)\) where \(k = \frac{1}{3}\).

Step2: Calculate new coordinates of \(S\)

For point \(S(0,-3)\), \(x = 0\), \(y=-3\), and \(k=\frac{1}{3}\). Then \(x'=k\times x=\frac{1}{3}\times0 = 0\) and \(y'=k\times y=\frac{1}{3}\times(-3)=-1\). So the new - coordinate of \(S\) is \((0,-1)\).

Step3: Calculate new coordinates of \(T\)

For point \(T(0,9)\), \(x = 0\), \(y = 9\), and \(k=\frac{1}{3}\). Then \(x'=k\times x=\frac{1}{3}\times0 = 0\) and \(y'=k\times y=\frac{1}{3}\times9 = 3\). So the new - coordinate of \(T\) is \((0,3)\).

Step4: Calculate new coordinates of \(U\)

For point \(U(3,9)\), \(x = 3\), \(y = 9\), and \(k=\frac{1}{3}\). Then \(x'=k\times x=\frac{1}{3}\times3 = 1\) and \(y'=k\times y=\frac{1}{3}\times9 = 3\). So the new - coordinate of \(U\) is \((1,3)\).

Step5: Calculate new coordinates of \(V\)

For point \(V(3,-9)\), \(x = 3\), \(y=-9\), and \(k=\frac{1}{3}\). Then \(x'=k\times x=\frac{1}{3}\times3 = 1\) and \(y'=k\times y=\frac{1}{3}\times(-9)=-3\). So the new - coordinate of \(V\) is \((1,-3)\).

Answer:

The new coordinates of the vertices are \(S(0,-1)\), \(T(0,3)\), \(U(1,3)\), \(V(1,-3)\)