Question

the figure below is dilated by a factor of $\frac{1}{4}$ centered at the origin. plot the resulting image. click twice to plot a segment. click a segment to delete it.

Explanation:

Step1: Identify the coordinates of the vertices

Let's assume the coordinates of points \(A\), \(B\), and \(C\) are \((x_A,y_A)\), \((x_B,y_B)\), and \((x_C,y_C)\) respectively. From the graph, if \(A=(6, - 8)\), \(B=(4,8)\), \(C=(-8,4)\)

Step2: Apply the dilation formula

The formula for dilation centered at the origin with a scale - factor \(k\) is \((x',y')=k(x,y)\). Here \(k = \frac{1}{4}\).
For point \(A\):
\(x'_A=\frac{1}{4}x_A=\frac{1}{4}\times6=\frac{3}{2}\)
\(y'_A=\frac{1}{4}y_A=\frac{1}{4}\times(-8)= - 2\)
So the new coordinates of \(A'\) are \((\frac{3}{2},-2)\)
For point \(B\):
\(x'_B=\frac{1}{4}x_B=\frac{1}{4}\times4 = 1\)
\(y'_B=\frac{1}{4}y_B=\frac{1}{4}\times8 = 2\)
So the new coordinates of \(B'\) are \((1,2)\)
For point \(C\):
\(x'_C=\frac{1}{4}x_C=\frac{1}{4}\times(-8)=-2\)
\(y'_C=\frac{1}{4}y_C=\frac{1}{4}\times4 = 1\)
So the new coordinates of \(C'\) are \((-2,1)\)

Step3: Plot the new points

Plot the points \(A'(\frac{3}{2},-2)\), \(B'(1,2)\) and \(C'(-2,1)\) on the coordinate - plane and connect them to form the dilated triangle.

Answer:

Plot the points \(A'(\frac{3}{2},-2)\), \(B'(1,2)\) and \(C'(-2,1)\) and connect them to get the dilated figure.