A line segment goes from $(1, 2)$ $(1 ,2 )$ to $(4, 7)$ $(4 ,7 )$ . The line segment is reflected across $x = 6$ $x=6$ , reflected across $y = - 1$ $y=-1$ , and then dilated about $(1, 1)$ $(1 ,1 )$ by a factor of $2$ $2$ . How far are the new endpoints from the origin?

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A line segment goes from $(1, 2)$ $(1 ,2 )$ to $(4, 7)$ $(4 ,7 )$ . The line segment is reflected across $x = 6$ $x=6$ , reflected across $y = - 1$ $y=-1$ , and then dilated about $(1, 1)$ $(1 ,1 )$ by a factor of $2$ $2$ . How far are the new endpoints from the origin?

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Answer 1 · 2016-07-07T03:37:31

Original segment $A_{0} B_{0}$ $A_0B_0$ , where $A_{0} = (1, 2), B_{0} = (4, 7)$ $A_0=(1,2), B_0=(4,7)$ ,
is transformed into $A B$ $AB$ , where $A = (21, - 9), B = (15, - 19)$ $A=(21,-9), B=(15,-19)$ .
The distances from the origin to the new endpoints are
$d_{A} \approx 22.8$ $d_A ~~22.8$
$d_{B} \approx 24.2$ $d_B ~~24.2$

Explanation:

Reflection of a point with coordinates $(a_{0}, b_{0})$ $(a_0,b_0)$ relative to a line $x = 6$ $x=6$ (vertical line intersecting X-axis at coordinate $x = 6$ $x=6$ ) will be horizontally shifted into a new X-coordinate obtained by adding to an X-coordinate of the axis of symmetry ( $x = 6$ $x=6$ ) the distance from it of the original X-coordinates ( $6 - a_{0}$ $6-a_0$ ).
Y-coordinate remains the same in this transformation.
So, new coordinates are:
$(a_{1}, b_{1}) = (6 + (6 - a_{0}), b_{0}) = (12 - a_{0}, b_{0})$ $(a_1,b_1) = (6+(6-a_0),b_0)=(12-a_0,b_0)$
Reflection of a point with coordinates $(a_{1}, b_{1})$ $(a_1,b_1)$ relative to a line $y = - 1$ $y=-1$ (horizontal line intersecting Y-axis at coordinate $y = - 1$ $y=-1$ ) will be vertically shifted into a new Y-coordinate obtained by adding to an Y-coordinate of the axis of symmetry ( $y = - 1$ $y=-1$ ) the distance from it of the original Y-coordinates ( $- 1 - b_{1}$ $-1-b_1$ ).
X-coordinate remains the same in this transformation.
So, new coordinates are:
$(a_{2}, b_{2}) = (a_{1}, - 1 + (- 1 - b_{1})) =$ $(a_2,b_2) = (a_1,-1+(-1-b_1))=$
$= (a_{1}, - 2 - b_{1}) = (12 - a_{0}, - 2 - b_{0})$ $= (a_1,-2-b_1)=(12-a_0,-2-b_0)$
Dilation about a center point $(1, 1)$ $(1,1)$ by a factor of $2$ $2$ will transform a point $(a_{2}, b_{2})$ $(a_2,b_2)$ into
$(a_{3}, b_{3}) = (1 + 2 (a_{2} - 1), 1 + 2 (b_{2} - 1)) =$ $(a_3,b_3) = (1+2(a_2-1),1+2(b_2-1)) =$
$= (1 + 2 (12 - a_{0} - 1), 1 + 2 (- 2 - b_{0} - 1)) =$ $= (1+2(12-a_0-1),1+2(-2-b_0-1)) =$
$= (23 - 2 a_{0}, - 5 - 2 b_{0})$ $= (23-2a_0, -5-2b_0)$
Using this formula for both ends of our original segment $A B$ $AB$ , where $A (1, 2)$ $A(1,2)$ and $B = (4, 7)$ $B=(4,7)$ :
4.1. $(a_{0} = 1, b_{0} = 2)$ $(a_0=1, b_0=2)$
$\to$ $rarr$ $(a_{3} = 23 - 2 \cdot 1, b_{3} = - 5 - 2 \cdot 2) =$ $(a_3=23-2*1, b_3=-5-2*2) =$
$= (21, - 9)$ $= (21, -9)$
4.2. $(a_{0} = 4, b_{0} = 7)$ $(a_0=4, b_0=7)$
$\to$ $rarr$ $(a_{3} = 23 - 2 \cdot 4, b_{3} = - 5 - 2 \cdot 7) =$ $(a_3=23-2*4, b_3=-5-2*7) =$
$= (15, - 19)$ $= (15, -19)$
The distance of each end of a new segment from the origin are
$d_{A} = \sqrt{{(21)}^{2} + {(- 9)}^{2}} = \sqrt{552} \approx 22.8$ $d_A = sqrt((21)^2+(-9)^2) = sqrt(552) ~~22.8$
$d_{B} = \sqrt{{(15)}^{2} + {(- 19)}^{2}} = \sqrt{586} \approx 24.2$ $d_B = sqrt((15)^2+(-19)^2) = sqrt(586) ~~24.2$

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A line segment goes from $(1, 2)$ $(1 ,2 )$ to $(4, 7)$ $(4 ,7 )$ . The line segment is reflected across $x = 6$ $x=6$ , reflected across $y = - 1$ $y=-1$ , and then dilated about $(1, 1)$ $(1 ,1 )$ by a factor of $2$ $2$ . How far are the new endpoints from the origin?

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A line segment goes from $(1, 2)$ $(1 ,2 )$ to $(4, 7)$ $(4 ,7 )$ . The line segment is reflected across $x = 6$ $x=6$ , reflected across $y = - 1$ $y=-1$ , and then dilated about $(1, 1)$ $(1 ,1 )$ by a factor of $2$ $2$ . How far are the new endpoints from the origin?