A line segment has endpoints at (8 , 4) and (1 , 2). If the line segment is rotated about the origin by (3pi)/2 , translated vertically by 4, and reflected about the x-axis, what will the line segment's new endpoints be?

Feb 26, 2018

Summarizing
transformed coordinates are
from
${x}_{11} = 8 , {y}_{11} = 4$
${x}_{21} = 1 , {y}_{21} = 2$
to
${x}_{14} , {y}_{14} = 4 , 4$
${x}_{24} , {y}_{24} = 2 , - 3$

Explanation:

${x}_{11} = 8 , {y}_{11} = 4$

${x}_{21} = 1 , {y}_{21} = 2$

${r}_{11} = \sqrt{{x}_{11}^{2} + {y}_{11}^{2}} = \sqrt{{8}^{2} + {4}^{2}} = \sqrt{64 + 16} = \sqrt{80}$

${\theta}_{11} = {\tan}^{-} 1 \left({y}_{11} / {x}_{11}\right) = {\tan}^{-} 1 \left(\frac{4}{2}\right) = {26.565}^{\circ}$

${r}_{21} = \sqrt{{x}_{21}^{2} + {y}_{21}^{2}} = \sqrt{{1}^{2} + {2}^{2}} = \sqrt{1 + 4} = \sqrt{5}$

${\theta}_{21} = {\tan}^{-} 1 \left({y}_{21} / {x}_{21}\right) = {\tan}^{-} 1 \left(\frac{2}{1}\right) = {63.435}^{\circ} ^ \circ$

Rotation is

$\alpha = \frac{3 \pi}{2} = {270}^{\circ}$

${r}_{12} = {r}_{11}$

${\theta}_{12} = {\theta}_{11} + {270}^{\circ}$

${r}_{12} = \sqrt{80}$

${\theta}_{12} = {26.565}^{\circ} + {270}^{\circ} = {296.565}^{\circ}$

${r}_{22} = {r}_{21}$

${\theta}_{22} = {\theta}_{21} + {270}^{\circ}$

${r}_{22} = \sqrt{5}$

${\theta}_{22} = {63.435}^{\circ} + {270}^{\circ} = {333.435}^{\circ}$

$\left({x}_{12} , {y}_{12}\right) = \left({r}_{12} \cos {\theta}_{12} , {r}_{12} \sin {\theta}_{12}\right)$
$= \sqrt{80} \cos {296.565}^{\circ} , \sqrt{80} \sin {296.565}^{\circ} = \equiv \left(4 , - 8\right)$

${x}_{12} = 4$
${y}_{12} = - 8$

$\left({x}_{22} , {y}_{22}\right) = \left({r}_{22} \cos {\theta}_{22} , {r}_{22} \sin {\theta}_{22}\right)$
$= \sqrt{5} \cos {333.435}^{\circ} , \sqrt{5} \sin {333.435}^{\circ} = \equiv \left(2 , - 1\right)$
${x}_{22} = 2$
${y}_{22} = - 1$

translation is +4

${x}_{13} = {x}_{12}$
${y}_{13} = {y}_{12} + 4$

${x}_{13} = 4$
${y}_{13} = - 8 + 4 = - 4$

${x}_{23} = {x}_{22}$
${y}_{23} = {y}_{22} + 4$

${x}_{23} = 2$
${y}_{23} = - 1 + 4 = 3$

${x}_{14} = {x}_{13}$
${y}_{14} = - {y}_{12}$

${x}_{14} = 4$
${y}_{14} = 4$

${x}_{24} = {x}_{23}$
${y}_{23} = - {y}_{22}$

${x}_{23} = 2$
${y}_{23} = - 3$

Summarizing

${x}_{11} , {y}_{11} = 8 , 4$
${x}_{21} , {y}_{21} = 1 , 2$
Rotation by $\frac{3 \pi}{2}$
${x}_{12} , {y}_{12} = 4 , - 8$
${x}_{22} , {y}_{22} = 2 , - 1$
translated bertically by 4
${x}_{13} , {y}_{13} = 4 , - 4$
${x}_{23} , {y}_{23} = 2 , 3$
${x}_{14} , {y}_{14} = 4 , 4$
${x}_{24} , {y}_{24} = 2 , - 3$