Question

dilate figure abcd by a scale factor of \\(\frac{1}{2}\\) with the center of dilation at the origin. find the coordinates of the vertices of the dilated image. figure abcd: a(4,4), b(6,2), c(6,6), d(4,8); figure abcd: a(2,2),? (for b), with options b(1,3), b(3,1), b(6,4), b(12,8). there is a coordinate grid with figure abcd plotted: a at (4,4), b at (6,2), c at (6,6), d at (4,8).

Explanation:

Step1: Recall dilation rule

To dilate a point \((x,y)\) with center at the origin and scale factor \(k\), the new coordinates are \((kx,ky)\). Here, \(k = \frac{1}{2}\).

Step2: Apply dilation to point B

For point \(B(6,2)\), multiply each coordinate by \(\frac{1}{2}\). So, \(x\)-coordinate: \(6\times\frac{1}{2}=3\), \(y\)-coordinate: \(2\times\frac{1}{2}=1\)? Wait, no, wait the first point \(A(4,4)\) becomes \(A'(2,2)\) which is \(4\times\frac{1}{2}=2\), \(4\times\frac{1}{2}=2\). Wait, maybe I misread. Wait, let's check the graph. Wait, the coordinates of B: looking at the graph, B is at (6,2)? Wait, no, in the graph, B is at (6,2)? Wait, the table has B(6,2), but the graph: A is at (4,4), D at (4,8), C at (6,6), B at (6,2). So to dilate by scale factor \(\frac{1}{2}\) from origin, we multiply each coordinate by \(\frac{1}{2}\). So for B(6,2), new x: \(6\times\frac{1}{2}=3\), new y: \(2\times\frac{1}{2}=1\)? But wait the options? Wait no, wait maybe I made a mistake. Wait the first point A(4,4) becomes (2,2), which is correct. So for B(6,2), applying scale factor \(\frac{1}{2}\), we get (6(1/2), 2(1/2)) = (3,1). Wait, looking at the dropdown, one of the options is B'(3,1). So that's the answer.

Wait, let's recheck: dilation with scale factor \(k\) from origin: \((x,y)\to(kx,ky)\). So for B(6,2), \(k=\frac{1}{2}\), so \(x' = 6\times\frac{1}{2}=3\), \(y' = 2\times\frac{1}{2}=1\). So B' is (3,1).

Answer:

\(B'(3,1)\)