Question

give the new coordinates for dilating trapezoid defg with vertices d(-5, 15), e(10, 10), f(10, 5), and g(-5, 0): k = \frac{1}{5}
there is no picture for this question.
write the numerical answer (ie if you get 2 for your answer, type \2\ not \two\)
d(
type your answer...
,
type your answer...
)
e(
type your answer...
,
type your answer...
)
f(
type your answer...
,
type your answer...
)
g(
type your answer...
,
type your answer...
)

Explanation:

Step1: Recall dilation rule

To dilate a point \((x,y)\) with a scale factor \(k\), the new coordinates \((x',y')\) are given by \(x' = k\times x\) and \(y'=k\times y\).

Step2: Dilate point D(-5, 15)

For point \(D(-5,15)\) and \(k = \frac{1}{5}\):
\(x'=\frac{1}{5}\times(-5)= - 1\)
\(y'=\frac{1}{5}\times15 = 3\)
So \(D'(-1,3)\)

Step3: Dilate point E(10, 10)

For point \(E(10,10)\) and \(k=\frac{1}{5}\):
\(x'=\frac{1}{5}\times10 = 2\)
\(y'=\frac{1}{5}\times10=2\)
So \(E'(2,2)\)

Step4: Dilate point F(10, 5)

For point \(F(10,5)\) and \(k = \frac{1}{5}\):
\(x'=\frac{1}{5}\times10 = 2\)
\(y'=\frac{1}{5}\times5 = 1\)
So \(F'(2,1)\)

Step5: Dilate point G(-5, 0)

For point \(G(-5,0)\) and \(k=\frac{1}{5}\):
\(x'=\frac{1}{5}\times(-5)=-1\)
\(y'=\frac{1}{5}\times0 = 0\)
So \(G'(-1,0)\)

Answer:

\(D'(-1, 3)\)
\(E'(2, 2)\)
\(F'(2, 1)\)
\(G'(-1, 0)\)