Question

A sequence of translations maps ∆ABC to ∆A’B’C’.
∆ABC has vertices A(-3, 4), B(-3, 0), C(-1, 3)
The coordinates for A’ are (3, -4)


(a) What is the coordinate rule that maps ∆ABC onto ∆A’B’C’?


(b) What are the coordinates for B’ and C’?

Answer 1

Answer:

(a) Co-ordinate rule is $x'=x+6$ and $y'=y-8$

(b) Co-ordinates of B' and C' are $(3,-8)$ and $(5,-5)$ respectively.

Step-by-step explanation:

(a)

Here, the co-ordinates of A$(-3,4)$ are translated to A'$(3,-4)$.

For the co-ordinates A and A',  $x'-x=3-(-3)=3+3=6$ and $y'-y=-4-4=-8$

So, x value of A has shifted to right by 6 units and y value of A has shifted 8 units down.

Hence, the co-ordinate rule that maps ΔABC onto ΔA'B'C' is:

$x'=x+6$ and $y'=y-8$.

(b)

Using the co-ordinate rule, we can find the co-ordinates of B' and C'.

For B, $x=-3$ and $y=0$.

So, $x'$ of B' is $x'=x+6=-3+6=3$

And, $y'$ of B' is $y'=y-8=0-8=-8$.

Therefore, co-ordinates of B' are $(3,-8)$.

For C, $x=-1$ and $y=3$.

So, $x'$ of C' is $x'=x+6=-1+6=5$

And, $y'$ of C' is $y'=y-8=3-8=-5$.

Therefore, co-ordinates of C' are $(5,-5)$.