Using displacement vector (((−4)),((−2)) ), what is the image of ((6),(3) ) when translated?


Question and Answers Forum

All Questions Topic List
Vector Questions
Previous in All Question Next in All Question
Previous in Vector Next in Vector
Question Number 113554 by Aina Samuel Temidayo last updated on 14/Sep/20

$Using displacement vector (\begin{matrix} - 4 \\ - 2 \end{matrix}),$ $what is the image of (\begin{matrix} 6 \\ 3 \end{matrix}) when$ $translated ?$
	Answered by bemath last updated on 14/Sep/20

	$s a y i m a g e o f (\begin{matrix} 6 \\ 3 \end{matrix}) i s (\begin{matrix} x \\ y \end{matrix})$ $\Leftrightarrow (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 6 \\ 3 \end{matrix}) + (\begin{matrix} - 4 \\ - 2 \end{matrix}) = (\begin{matrix} 2 \\ 1 \end{matrix})$
	Commented by Aina Samuel Temidayo last updated on 14/Sep/20

	$Why did you add them together$ $please ? Explain in details . What is a$ $displacement vector ?$
	Answered by mathmax by abdo last updated on 14/Sep/20
	$let A ((6),(3) ) and A^′ ((x),(y) )=t_u^⌣ (A) ⇒AA^→ ′ =u^→ we have u^→ (((−4)),((−2)) ) ⇒ (((x−6)),((y−3)) ) = (((−4)),((−2)) ) ⇒ { ((x−6 =−4)),((y−3 =−2)) :} ⇒ { ((x=2)),((y =1)) :} ⇒A^′ ((2),(1) )$
	$let A (\begin{matrix} 6 \\ 3 \end{matrix}) and A^{^{'}} (\begin{matrix} x \\ y \end{matrix}) = t_{\overset{⌣}{u}} (A) \Rightarrow A {\vec{A}}^{'} = \vec{u} we have \vec{u} (\begin{matrix} - 4 \\ - 2 \end{matrix})$ $\Rightarrow (\begin{matrix} x - 6 \\ y - 3 \end{matrix}) = (\begin{matrix} - 4 \\ - 2 \end{matrix}) \Rightarrow {\begin{cases} x - 6 = - 4 \\ y - 3 = - 2 \end{cases}$ $\Rightarrow {\begin{cases} x = 2 \\ y = 1 \end{cases}$ $\Rightarrow A^{^{'}} (\begin{matrix} 2 \\ 1 \end{matrix})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum