Question

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

Explanation:

1. First, identify the original coordinates of the vertices:

• Assume the coordinates of point \(B\) are \((-3,-9)\), the coordinates of point \(C\) are \((-3,6)\), and assume there is a point on the line - segment intersecting the \(y\) - axis at \((0,2)\) (let's call it \(A\)), and the coordinates of point \(D\) are \((9, - 9)\).

1. Recall the rule for dilation centered at the origin:

• If a point \((x,y)\) is dilated with a scale - factor \(k\) centered at the origin, the new coordinates \((x',y')\) are given by \((x',y')=(k x,k y)\). Here, \(k = \frac{1}{3}\).
• For point \(A(0,2)\):
• \(x'=\frac{1}{3}\times0 = 0\) and \(y'=\frac{1}{3}\times2=\frac{2}{3}\). So the new coordinates of \(A\) are \((0,\frac{2}{3})\).
• For point \(B(-3,-9)\):
• \(x'=\frac{1}{3}\times(-3)=-1\) and \(y'=\frac{1}{3}\times(-9)= - 3\). So the new coordinates of \(B\) are \((-1,-3)\).
• For point \(C(-3,6)\):
• \(x'=\frac{1}{3}\times(-3)=-1\) and \(y'=\frac{1}{3}\times6 = 2\). So the new coordinates of \(C\) are \((-1,2)\).
• For point \(D(9,-9)\):
• \(x'=\frac{1}{3}\times9 = 3\) and \(y'=\frac{1}{3}\times(-9)=-3\). So the new coordinates of \(D\) are \((3,-3)\).

Step1: Identify original coordinates

Assume \(A(0,2)\), \(B(-3,-9)\), \(C(-3,6)\), \(D(9,-9)\)

Step2: Apply dilation formula

For \(A\): \((x',y')=(\frac{1}{3}\times0,\frac{1}{3}\times2)=(0,\frac{2}{3})\)
For \(B\): \((x',y')=(\frac{1}{3}\times(-3),\frac{1}{3}\times(-9))=(-1,-3)\)
For \(C\): \((x',y')=(\frac{1}{3}\times(-3),\frac{1}{3}\times6)=(-1,2)\)
For \(D\): \((x',y')=(\frac{1}{3}\times9,\frac{1}{3}\times(-9))=(3,-3)\)

Answer:

The new coordinates of \(A\) are \((0,\frac{2}{3})\), of \(B\) are \((-1,-3)\), of \(C\) are \((-1,2)\), of \(D\) are \((3,-3)\)