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

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 (\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})