Question

1. trapezoid defg with vertices d(-5,15), e(10,10), f(10,5), and g(-5,0): k = 1/5 d (__, ) e (, ) f (, ) g (, __)

Explanation:

Step1: Apply scaling to D(-5,15)

To find the image of a point \((x,y)\) under a scale factor \(k\), we use the formula \((kx, ky)\). For point \(D(-5,15)\) and \(k = \frac{1}{5}\), we calculate \(x\)-coordinate: \(-5\times\frac{1}{5}=-1\), \(y\)-coordinate: \(15\times\frac{1}{5}=3\). So \(D'(-1, 3)\).

Step2: Apply scaling to E(10,10)

For point \(E(10,10)\), \(x\)-coordinate: \(10\times\frac{1}{5}=2\), \(y\)-coordinate: \(10\times\frac{1}{5}=2\). So \(E'(2, 2)\).

Step3: Apply scaling to F(10,5)

For point \(F(10,5)\), \(x\)-coordinate: \(10\times\frac{1}{5}=2\), \(y\)-coordinate: \(5\times\frac{1}{5}=1\). So \(F'(2, 1)\).

Step4: Apply scaling to G(-5,0)

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

Answer:

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