Question

The endpoints of (line segment )AB are A(2, 2) and B(3, 8). (line segment ) AB is dilated by a scale factor of 3.5 with the origin as the center of dilation to give image (line segment ) A'B' . What are the slope (m) and length of (line segment ) A'B'? Use the distance formula to help you decide

Answer 1

The slope:
m = ( y2 - y1 ) / ( x2 - x1 ) = ( 8  2 ) / ( 3 - 2 ) = 6 / 1 
m = 6  ( we have the same slope for AB and A`B` )
AB = √[( 3 - 2 )² + ( 8 - 2 )²] = √37
A`B` = 3.5 √37 = 21.29 

Answer 2

Answer: The slope of A'B' is 6 and its length is 21.29 units .

Step-by-step explanation:  Given that the end-points of the line segment AB are (2, 2) and B(3, 8). AB is dilated by a scale factor of 3.5 with the centre of dilation as origin o form the line segment A'B'.

We are to find the slope (m) and length of line segment A'B'.

DISTANCE FORMULA:  The distance between two points (a, b) and (c, d) is given by

$D=\sqrt{(c-a)^2+(d-b)^2}.$

So, the length of the line segment AB is

$L_{AB}=\sqrt{(3-2)^2+(8-2)^2}=\sqrt{1+36}=\sqrt{37}.$

Since the line segment AB is dilated by a scale factor of 3,5, so the length of A'B' will be

$L_{A'B'}=3.5\times L_{AB}=3.5\times \sqrt{37}=3.5\sqrt{37}=2.29~\textup{units}.$

The centre of dilation is origin, so the line segment A'B' is an extension of AB, and the points (2, 2) and (3, 8) will also lie on A'B'.

Therefore, the slope of A'B' will be

$m=\dfrac{8-2}{3-2}=6.$

Thus, the slope of A'B' is 6 and its length is 21.29 units.