Question

The vertices of ∆ABC are A(7,3),B(−8,6),C(0,−5). If ∆ABC is translated along the vector <−9,5>, what are the coordinates of the vertices of ∆A'B'C' ?

Answer 1

Answer:

The coordinates of the vertices of $\bigtriangleup A'B'C'$ are $A'=(-2,8)$, $B'=(-17,-4)$ and $C'=(-9,0)$.

Step-by-step explanation:

Let be $A =(7,3)$, $B = (-8,6)$ and $C = (0,-5)$ the vertices of the original triangle, as well as $\vec t = \left$, the coordinates of the vertices of the new triangle are, respectively:

$A' = A + \vec t$

$A' = (7,3)+(-9,5)$

$A' = (7-9,3+5)$

$A'=(-2,8)$

$B' = B + \vec t$

$B' = (-8,6)+(-9,5)$

$B' = (-8-9,-9+5)$

$B'=(-17,-4)$

$C' = C + \vec t$

$C' = (0,-5)+(-9,5)$

$C' = (0-9,-5+5)$

$C'=(-9,0)$

The coordinates of the vertices of $\bigtriangleup A'B'C'$ are $A'=(-2,8)$, $B'=(-17,-4)$ and $C'=(-9,0)$.