A line segment goes from $(1, 5)$ $(1 ,5 )$ to $(7, 6)$ $(7 ,6 )$ . The line segment is dilated about $(1, 2)$ $(1 ,2 )$ by a factor of $3$ $3$ . Then the line segment is reflected across the lines $x = 4$ $x = 4$ and $y = - 2$ $y=-2$ , in that order. How far are the new endpoints form the origin?

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A line segment goes from $(1, 5)$ $(1 ,5 )$ to $(7, 6)$ $(7 ,6 )$ . The line segment is dilated about $(1, 2)$ $(1 ,2 )$ by a factor of $3$ $3$ . Then the line segment is reflected across the lines $x = 4$ $x = 4$ and $y = - 2$ $y=-2$ , in that order. How far are the new endpoints form the origin?

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Answer 1 · 2018-05-30T01:07:42

new endpoints: $A (- 11, - 18), B (7, - 15)$ $A(-11,-18), B(7, -15)$
$d_{A} = \sqrt{445} \approx 21.1; d_{B} = \sqrt{274} \approx 16.6$ $d_A = sqrt(445) ~~21.1; d_B = sqrt(274)~~16.6$

Explanation:

Given: $A (1, 5), B (7, 6)$ $A(1, 5), B(7,6)$ . Line $¯ ¯¯¯¯ ¯ A B$ $bar (AB)$ is dilated about $(1, 2)$ $(1,2)$ by a factor $k = 3$ $k = 3$ . Then the line segment is reflected across $x = 4$ $x = 4$ and $y = - 2$ $y = -2$ .

Dilated about $(1, 2)$ $(1,2)$ . Find the distance to each endpoint from the dilated point:

$d_{A \to (1, 2)} = \sqrt{{(5 - 2)}^{2} + {(1 - 1)}^{2}} = \sqrt{9} = 3$ $d_(A to (1,2)) = sqrt((5-2)^2 + (1-1)^2) = sqrt(9) = 3$

$d_{B \to (1, 2)} = \sqrt{{(6 - 2)}^{2} + {(7 - 1)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13}$ $d_(B to (1,2)) = sqrt((6-2)^2 + (7-1)^2) = sqrt(16+36) = sqrt(52) = 2sqrt(13)$

Triple the distance (dilate the length to the point $(1, 2)) :$ $(1, 2)):$
$3 (3) = 9; 3 \cdot \sqrt{52} = 3 \sqrt{52} = 6 \sqrt{13}$ $3(3) = 9; " " 3*sqrt(52) = 3sqrt(52) = 6sqrt(13)$

Since $(1, 2)$ $(1,2)$ & $(1, 5)$ $(1,5)$ have the same $x$ $x$ value, the dilated point has the same $x$ $x$ value:

$" "A'(1, 2+9) = A'(1, 11)$
The 2 is added from the dilated point's $y$ value.

Use proportions to find the $B'(x_B, y_B)$ coordinates:

$(2sqrt(13))/(6sqrt(13)) = 6/x; " "2sqrt(13)x = 36 sqrt(13); " "x = (36 sqrt(13))/(2 sqrt(13)) = 18$

$x_B = 1 + 18 = 19$
The 1 is added from the dilated point's $x$ value.

$(2sqrt(13))/(6sqrt(13)) = 4/y; " "2sqrt(13)y = 24 sqrt(13); " "y = (24 sqrt(13))/(2 sqrt(13)) = 12$

$y_B = 2 + 12 = 14$
The 2 is added from the dilated point's $y$ value.

Dilated line segment points: $" "A'(1, 11); " "B'(19, 14)$

Reflected about the $x = 4$ line:

$x$ distance from $x = 4$ to $x_(A') = 3$
$x$ distance from $x = 4$ to $x_(B') = 15$

Since the points are on opposite sides of $x = 4$ there are two coordinate rules:

coordinate rule of reflected point A' : $(x, y) -> ( 4+x"-distance", y)$
coordinate rule of reflected point B' : $(x, y) -> (4 - x"-distance", y)$

$A''(4+3, 11) = (7, 11)$
$B''(4 - 15, 14) = (-11, 14)$

Reflected about $x = 4$ line segment points: $A'(7, 11); " "B'(-11, 14)$

Reflected about the $y = -2$ line:

$y$ distance from $y = 2$ to $y_(A'') = 13$
$y$ distance from $y = 2$ to $y_(B'') = 16$

coordinate rule of reflected points : $(x, y) -> ( x, -2 - y "-distance")$

$A'''(7, -2 - 13) = (7, -15)$
$B'''(-11, -2 - 16) = (-11, -18)$

Reflected about $y = -2$ line segment points: $A'(7, -15); " "B'(-11, -18)$

distance of each endpoint from the origin:

$d_(A''' to (0,0))= sqrt(7^2 + (-15)^2) = sqrt(274)~~16.6$

$d_(B''' to (0,0)) = sqrt((-11)^2 + (-18)^2) = sqrt(445)~~21.1$

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