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The-point-4-1-undergoes-the-following-two-successive-transformations-i-reflection-about-the-line-y-x-ii-translation-through-a-distance-2-units-along-the-positive-x-axis-then-find-the-final-coord




Question Number 77001 by vishalbhardwaj last updated on 02/Jan/20
The point (4,1) undergoes the following  two successive transformations  (i) reflection about the line y=x  (ii) translation through a distance  2 units along the positive x axis  then find the final coordinates  of the point ????
$$\mathrm{The}\:\mathrm{point}\:\left(\mathrm{4},\mathrm{1}\right)\:\mathrm{undergoes}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{two}\:\mathrm{successive}\:\mathrm{transformations} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{reflection}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{x} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{translation}\:\mathrm{through}\:\mathrm{a}\:\mathrm{distance} \\ $$$$\mathrm{2}\:\mathrm{units}\:\mathrm{along}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{x}\:\mathrm{axis} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{final}\:\mathrm{coordinates} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:???? \\ $$
Answered by MJS last updated on 02/Jan/20
reflection about y=x just exchanges the  values of x and y  (i) →  ((1),(4) )  (ii) →  ((3),(4) )
$$\mathrm{reflection}\:\mathrm{about}\:{y}={x}\:\mathrm{just}\:\mathrm{exchanges}\:\mathrm{the} \\ $$$$\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{and}\:{y} \\ $$$$\left(\mathrm{i}\right)\:\rightarrow\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{4}}\end{pmatrix} \\ $$$$\left(\mathrm{ii}\right)\:\rightarrow\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{4}}\end{pmatrix} \\ $$
Commented by vishalbhardwaj last updated on 03/Jan/20
I did not get it sir, please explain...
$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{get}\:\mathrm{it}\:\mathrm{sir},\:\mathrm{please}\:\mathrm{explain}… \\ $$
Commented by MJS last updated on 03/Jan/20
just draw it. P= ((x),(y) )  reflected about y=x  ⇒ P_1 = ((y),(x) ). translation along the positive  x−axis = slide to the right parallel to x−axis  add distance to x−value P_2 = (((y+d)),(x) )
$$\mathrm{just}\:\mathrm{draw}\:\mathrm{it}.\:{P}=\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:\mathrm{reflected}\:\mathrm{about}\:{y}={x} \\ $$$$\Rightarrow\:{P}_{\mathrm{1}} =\begin{pmatrix}{{y}}\\{{x}}\end{pmatrix}.\:\mathrm{translation}\:\mathrm{along}\:\mathrm{the}\:\mathrm{positive} \\ $$$${x}−\mathrm{axis}\:=\:\mathrm{slide}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{parallel}\:\mathrm{to}\:{x}−\mathrm{axis} \\ $$$$\mathrm{add}\:{d}\mathrm{istance}\:\mathrm{to}\:{x}−\mathrm{value}\:{P}_{\mathrm{2}} =\begin{pmatrix}{{y}+{d}}\\{{x}}\end{pmatrix} \\ $$

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