Question

draw the triangle that results from the following sequence of transformations: 1. dilate by a factor of $\frac{1}{3}$ centered at the origin. 2. rotate 180° counterclockwise.

Explanation:

Step1: Dilate the vertices

Let the vertices of the original triangle be \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\). For a dilation centered at the origin with a scale - factor \(k = \frac{1}{3}\), the new vertices \(A'(x_1',y_1')\), \(B'(x_2',y_2')\), \(C'(x_3',y_3')\) are given by \(x_i'=\frac{1}{3}x_i\) and \(y_i'=\frac{1}{3}y_i\) for \(i = 1,2,3\).
If \(A=(4,6)\), then \(A'=(\frac{4}{3},2)\); if \(B=(8,3)\), then \(B'=(\frac{8}{3},1)\); if \(C=(5, - 3)\), then \(C'=(\frac{5}{3},-1)\).

Step2: Rotate the dilated vertices

The rule for a \(180^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-x,-y)\).
For \(A'=(\frac{4}{3},2)\), the rotated point \(A''=(-\frac{4}{3},-2)\).
For \(B'=(\frac{8}{3},1)\), the rotated point \(B''=(-\frac{8}{3},-1)\).
For \(C'=(\frac{5}{3},-1)\), the rotated point \(C''=(-\frac{5}{3},1)\).

Answer:

Plot the triangle with vertices \((-\frac{4}{3},-2)\), \((-\frac{8}{3},-1)\), \((-\frac{5}{3},1)\) on the coordinate - grid.