Question

4
The endpoints of AB are A(2, 2) and B(3,8). AB Is dilated by a scale factor of 3.5 with the origin as the center of dilation to give Image A'B'
What are the slope (m) and length of A'B' ? Use the distance formula to help you decide: d = JQxz - x1 ) +(y2-y).
OA. m = 6, A'B'=/37
OB. m=21, A'B'= 3.537
OC. m=6, A'B'= 3.5/37
OD.m=21, A'B'=/37
​

Answer 1

Given:

The end-points of the line of AB are A(2,2) and B(3,8).

Dilation/ Scale factor = 3.5

To Find:

We have to find the slope, m, and length of $A^{'}B^{'}$

Solution:

The slope of the line AB doesn't change.

  The equation of the slope, m =  $\frac{y_{2} -y_{1} }{x_{2}- x_{1} }$

                                                    = $\frac{8-2}{3-2}$

                                                    = 6

∴ The equation of the slope, m = 6

The scale factor in the dilation of a mathematical object determines how much larger or smaller is the object compared to the original size. When the value of the factor is greater than one then expansion occurs.

Here the scale factor is 3.5. Therefore, the length expands by the dilation factor.

The length of AB is given by the distance formula,

        $\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]$

Where

d is the distance

$\[\left( {{x_1},{y_1}} \right)\]$ are the coordinates of the first point

$\[\left( {{x_2},{y_2}} \right)\]$ are the coordinates of the second point

⇒

       $\[\begin{array}{l}d = \sqrt {{{\left( {3 - 2} \right)}^2} + {{\left( {8 - 2} \right)}^2}} \\\\d = \sqrt {37} \end{array}\]$

The length of $A^{'}B^{'}$ is

       $\[\begin{array}{l}{d^'} = 3 \cdot 5 \times \sqrt {37} \\\\{d^'} = 21.29\end{array}\]$

∴ The length of $A^{'}B^{'}$ is 21·29.