Reflect a triangle ABC, A(2,4), B(4,6) & C(2,6) about a line 2y - x - 4 = 0. Find out the new vertices of a triangle.

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written 8.2 years ago by

teamques10 ★ 69k

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2Y – X – 4 = 0

Y = 0.5X +2

tan⁡θ = m = 0.5

∴y=mx+c

θ=26.56°

Steps:-

Tr = Translate point ( 0, -2 ) to origin O ( 0, 0 )
R = Rotate the line about origin by .θ=-26.56°
Reflect the object about x-axis( Mx)
Inverse Rotate the line about origin by .θ=26.56°
$Tr^{-1}$ = Inverse Translation to original position a ( 0, 2)

$[T] = (Tr) (R) (Mx) (R)^{-1} (T)^{-1}$

$[ T ] = \begin{bmatrix}1&0&0\\0&1&0\\tx&ty&1\end{bmatrix} \times \begin{bmatrix}\cos⁡-26.56& \sin⁡ -26.56&0 \\ -\sin -26.56 & \cos⁡ -26.56 &0 \\ 0&0&1\end{bmatrix} \times \begin{bmatrix}1&0&0 \\ 0&-1&0 \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix}\cos⁡-26.56& \sin⁡ -26.56&0 \\ -\sin -26.56 & \cos⁡ -26.56 &0 \\ 0&0&1\end{bmatrix} \times \begin{bmatrix}1&0&0\\0&1&0\\tx&ty&1\end{bmatrix} \\ [ T ] = \begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&-2&1\end{bmatrix} \times \begin{bmatrix}\cos⁡-26.56& \sin⁡ -26.56&0 \\ -\sin -26.56 & \cos⁡ -26.56 &0 \\ 0&0&1\end{bmatrix} \times \begin{bmatrix}1&0&0 \\ 0&-1&0 \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix}\cos⁡-26.56& \sin⁡ -26.56&0 \\ -\sin -26.56 & \cos⁡ -26.56 &0 \\ 0&0&1\end{bmatrix} \times \begin{bmatrix}1&0&0 \\ 0&1&0 \\ 0&2&1\end{bmatrix} \\ [ T ] = \begin{bmatrix}0.6001&0.7995&0 \\ 0.7995&-0.6&0 \\ 0.894&0.2112&1\end{bmatrix} \\ \begin{bmatrix}A' \\ B' \\ C' \end{bmatrix} = \begin{bmatrix}2&4&1 \\ 4&6&1 \\ 2&6&1\end{bmatrix} \times \begin{bmatrix}0.6001&0.7995&0 \\ 0.7995&-0.6&0 \\ 0.894&0.2112&1\end{bmatrix} \\ \begin{bmatrix}A' \\ B' \\ C'\end{bmatrix}=\begin{bmatrix}2.8&2.4&1 \\ 5.6&2.8&1 \\ 4.4&1.2&1\end{bmatrix} \\ ∴A'=2.8 ,2.4 \\ ∴B'=5.6 ,2.8 \\ ∴C'=4.4 ,1.2$

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