QUESTION IMAGE
Question
- quadrilateral jklm is the image of quadrilateral jklm after a dilation. point k is the center of dilation. what is the scale factor of the dilation? a. 3 b. 2 c. 1/2 d. 1/3 6. triangle abc is dilated with a scale factor of 4 and the center of dilation is the origin. if the coordinates of the vertices of △abc are a(4, 8), b(12, 8), and c(16, 20), what are the coordinates of the vertices of the image after the dilation? a. a(16, 32), b(48, 32), c(64, 80) b. a(16, 32), b(48, 8), c(48, 80) c. a(8, 12), b(16, 12), c(20, 24) d. a(1, 2), b(3, 2), c(4, 5)
Step1: Recall dilation formula
For a dilation with center at the origin \((0,0)\) and scale - factor \(k\), if a point \(P(x,y)\) is dilated, the image \(P'(x',y')\) is given by \(x' = kx\) and \(y'=ky\).
Step2: Solve problem 5
Let's take a point, say \(M(2, - 2)\) and its image \(M'(1,-1)\). Using the dilation formula \(x'=kx\) and \(y' = ky\). Substituting \(x = 2,y=-2,x' = 1,y'=-1\) into \(x'=kx\), we get \(1=k\times2\), so \(k=\frac{1}{2}\). We can check with the \(y\) - coordinate: \(-1=k\times(-2)\), and when \(k = \frac{1}{2}\), \(-1=\frac{1}{2}\times(-2)\). So the scale - factor of the dilation in problem 5 is \(\frac{1}{2}\), and the answer is C.
Step3: Solve problem 6
Given \(A(4,8)\), \(B(12,8)\), \(C(16,20)\) and scale - factor \(k = 4\).
For point \(A\):
Using the dilation formula \(x'=kx\) and \(y' = ky\), \(x'=4\times4 = 16\) and \(y'=4\times8 = 32\), so \(A'(16,32)\).
For point \(B\):
\(x'=4\times12 = 48\) and \(y'=4\times8 = 32\), so \(B'(48,32)\).
For point \(C\):
\(x'=4\times16 = 64\) and \(y'=4\times20 = 80\), so \(C'(64,80)\). The answer is A.
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- C. \(\frac{1}{2}\)
- A. \(A'(16,32)\), \(B'(48,32)\), \(C'(64,80)\)