Using-displacement-vector-4-2-what-is-the-image-of-6-3-when-translated-

Question Number 113554 by Aina Samuel Temidayo last updated on 14/Sep/20

$$\mathrm{Using}\:\mathrm{displacement}\:\mathrm{vector}\:\begin{pmatrix}{−\mathrm{4}}\\{−\mathrm{2}}\end{pmatrix}, \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:\mathrm{when} \\ $$$$\mathrm{translated}? \\ $$

Answered by bemath last updated on 14/Sep/20

say image of ((6),(3) ) is ((x),(y) ) ⇔ ((x),(y) ) = ((6),(3) )+ (((−4)),((−2)) )= ((2),(1) )

$${say}\:{image}\:{of}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:{is}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{−\mathrm{4}}\\{−\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix} \\ $$

Commented by Aina Samuel Temidayo last updated on 14/Sep/20

Why did you add them together please? Explain in details. What is a displacement vector?

$$\mathrm{Why}\:\mathrm{did}\:\mathrm{you}\:\mathrm{add}\:\mathrm{them}\:\mathrm{together} \\ $$$$\mathrm{please}?\:\mathrm{Explain}\:\mathrm{in}\:\mathrm{details}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{displacement}\:\mathrm{vector}? \\ $$

Answered by mathmax by abdo last updated on 14/Sep/20

let A ((6),(3) ) and A^′ ((x),(y) )=t_u^⌣ (A) ⇒AA^→ ′ =u^→ we have u^→ (((−4)),((−2)) ) ⇒ (((x−6)),((y−3)) ) = (((−4)),((−2)) ) ⇒ { ((x−6 =−4)),((y−3 =−2)) :} ⇒ { ((x=2)),((y =1)) :} ⇒A^′ ((2),(1) )

$$\mathrm{let}\:\mathrm{A}\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:\:\:\mathrm{and}\:\mathrm{A}^{'} \begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}=\mathrm{t}_{\overset{\smile} {\mathrm{u}}} \left(\mathrm{A}\right)\:\Rightarrow\mathrm{A}\overset{\rightarrow} {\mathrm{A}}'\:=\overset{\rightarrow} {\mathrm{u}}\:\:\:\:\:\:\mathrm{we}\:\mathrm{have}\:\overset{\rightarrow} {\mathrm{u}}\begin{pmatrix}{−\mathrm{4}}\\{−\mathrm{2}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{x}−\mathrm{6}}\\{\mathrm{y}−\mathrm{3}}\end{pmatrix}\:=\begin{pmatrix}{−\mathrm{4}}\\{−\mathrm{2}}\end{pmatrix}\:\Rightarrow\:\begin{cases}{\mathrm{x}−\mathrm{6}\:=−\mathrm{4}}\\{\mathrm{y}−\mathrm{3}\:=−\mathrm{2}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{2}}\\{\mathrm{y}\:=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\mathrm{A}^{'} \begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix} \\ $$

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