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

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Answer 1 · 2018-04-12T03:10:20

Distance of the points from the origin after dilation and reflection :

$color(crimson)(bar(OA''') = 11.66, bar(OB''') = 20.62$

Given : $A(6,5), B(7,3)$

Now we have to find A', B' after rotation about point C (2,1) with a dilation factor of 3.

$A'(x, y) -> 3 * A(x,y) - C (x,y)$

$A'((x),(y)) = 3 * ((6), (5)) - ((2), (1)) = ((16),(14))$

Similarly, $B'((x),(y)) = 3 * ((7), (3)) - ((2), (1)) = ((19),(8))$

Now to find

Reflection Rules :

$color(crimson)("reflect over a line. ex: x=6, (2h-x, y)"$

$"Reflected across "x = 3$

$A'(x,y) -> A''(x,y) = ((6 - 16), (14)) = (-10, 14)$

$B'(x,y) -> B''(x,y) = ((2 * 3 - 19), (8)) = (-13,8)$

$"Reflected across " y = -4$

$color(crimson)("reflect over a line. ex: y= -3, (x, 2k-y)"$

$A''(x,y) -> A'''(x,y) = ((-10, 8-14) = (-10,6)$

$B''(x,y) -> B'''(x,y) =( -13, -8-8) = (-13, -16)$

$bar(OA''') = sqrt(-10^2 + 6^2) = 11.66$

$bar(OB''') = sqrt(-13^2 + -16^2) = 20.62$

A line segment goes from (6 ,5 )(6,5)(6 ,5 ) to (7 ,3 )(7,3)(7 ,3 ). The line segment is dilated about (2 ,1 )(2,1)(2 ,1 ) by a factor of 333. Then the line segment is reflected across the lines x = 3x=3x = 3 and y=-4y=−4y=-4, in that order. How far are the new endpoints form the origin?