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:

• From the graph, we can see that the coordinates of point \(B\) are \((- 9,9)\), the coordinates of point \(C\) are \((9,9)\), and assume the origin \(O(0,0)\) is also a vertex - like in a triangular - like shape formed by the lines and the origin (if we consider the dilation context). Let's also assume we can find the coordinates of point \(D\) which seems to be \((-9, - 9)\).
• The rule for dilation centered at the origin with a scale factor \(k\) is \((x,y)\to(kx,ky)\). Here \(k = \frac{1}{3}\).

1. Calculate the new coordinates of point \(B\):

• For point \(B(-9,9)\), using the dilation formula \((x,y)\to(kx,ky)\) with \(k=\frac{1}{3}\), we have \(x=-9\) and \(y = 9\).
• The new \(x\) - coordinate is \(x_{new}=k\times x=\frac{1}{3}\times(-9)=-3\).
• The new \(y\) - coordinate is \(y_{new}=k\times y=\frac{1}{3}\times9 = 3\). So the new coordinates of \(B\) are \((-3,3)\).

1. Calculate the new coordinates of point \(C\):

• For point \(C(9,9)\), with \(x = 9\) and \(y = 9\) and \(k=\frac{1}{3}\).
• The new \(x\) - coordinate is \(x_{new}=k\times x=\frac{1}{3}\times9 = 3\).
• The new \(y\) - coordinate is \(y_{new}=k\times y=\frac{1}{3}\times9 = 3\). So the new coordinates of \(C\) are \((3,3)\).

1. Calculate the new coordinates of point \(D\):

• For point \(D(-9,-9)\), with \(x=-9\) and \(y=-9\) and \(k = \frac{1}{3}\).
• The new \(x\) - coordinate is \(x_{new}=k\times x=\frac{1}{3}\times(-9)=-3\).
• The new \(y\) - coordinate is \(y_{new}=k\times y=\frac{1}{3}\times(-9)=-3\). So the new coordinates of \(D\) are \((-3,-3)\).

Answer:

The new coordinates of \(B\) are \((-3,3)\), the new coordinates of \(C\) are \((3,3)\), and the new coordinates of \(D\) are \((-3,-3)\).