Answer:
The slope is m=6 and length of line is [tex]d' =3.5\sqrt{37}[/tex]
Step-by-step explanation:
Given : The endpoints of line AB are A(2, 2) and B(3, 8). Line AB is dilated by a scale factor of 3.5 with the origin as the center of dilation to give image line A'B' .
To find : What are the slope (m) and length of line AB?
Solution :
The slope of a line does not change.
Slope formula is,
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
The slope of AB A(2, 2) and B(3, 8) is :
[tex]m=\frac{8-2}{3-2}\\\\m=\frac{6}{1}\\\\m=6[/tex]
Distance formula, [tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
The length of AB by distance formula,
[tex]d = \sqrt{(3 - 2)^2+(8-2)^2}\\d = \sqrt{(1)^2+(6)^2}\\d=\sqrt{1+36}\\d=\sqrt{37}[/tex]
Line AB is dilated by a scale factor of 3.5.
The length of A'B' is :
[tex]d' =3.5\times\sqrt{37}\\d' =3.5\sqrt{37}[/tex]
Therefore, The slope is m=6 and length of line is [tex]d' =3.5\sqrt{37}[/tex]
