Question

select the correct answer from each drop - down menu. rectangle abcd is dilated by a scale factor of \\( \frac { 1 } { 4 } \\) to form \\( a ^ { \prime } b ^ { \prime } c ^ { \prime } d ^ { \prime } \\). point t is the center of dilation and lies on line segment ab as shown. in \\( a ^ { \prime } b ^ { \prime } c ^ { \prime } d ^ { \prime } \\), line segment \\( a ^ { \prime } b ^ { \prime } \\) has a slope of 0 and a length of

Explanation:

Step1: Find the slope of \(A'B'\)

Dilation preserves the slope of a line segment. Since \(AB\) is a horizontal line (all points on \(AB\) have the same \(y -\)coordinate, \(y = 4\)), the slope of \(AB\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \(A=(1,4)\) and \(B=(6,4)\), then \(m=\frac{4 - 4}{6 - 1}=0\). So the slope of \(A'B'\) is \(0\) (because dilation is a similarity transformation that preserves parallelism, and horizontal lines remain horizontal).

Step2: Calculate the length of \(AB\)

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for \(A=(1,4)\) and \(B=(6,4)\). Then \(d_{AB}=\sqrt{(6 - 1)^2+(4 - 4)^2}=\sqrt{25+0}=5\).

Step3: Calculate the length of \(A'B'\)

The scale factor of dilation is \(k = \frac{1}{4}\). The length of a dilated segment \(d_{A'B'}=k\times d_{AB}\). Substitute \(k=\frac{1}{4}\) and \(d_{AB} = 5\) into the formula, we get \(d_{A'B'}=\frac{1}{4}\times5 = 1.25\)

Answer:

The slope of \(A'B'\) is \(0\) and the length of \(A'B'\) is \(1.25\)