Question

3. The coordinator of ΔABC are A(-4,5) B(1,4) and C(0,-3). Great the coordinates of A’, B’, and C’ when ΔABC Is under a translation 3 units to the right and 7 units down

Answer 1

The ΔABC has the following coordinates:

`\begin{gathered} A=(-4,5) \\ B=(1,4) \\ C=(0,-3) \end{gathered}`

The triangle undergoes a translation to the right and down.

If a point is shifted "a" units to the right, the transformation rule is given to be:

`(x,y)\to(x+a,y)`

If a point is shifted "b" units down, the transformation rule is given to be:

`(x,y)\to(x,y-b)`

Combining both rules, we have:

`(x,y)\to(x+a,y-b)`

If the original points are translated 3 units to the right and 7 units down, we will have:

`\begin{gathered} a=3 \\ b=7 \end{gathered}`

Hence, the rule becomes:

`(x,y)\to(x+3,y-7)`

Therefore, the new coordinates are:

`\begin{gathered} A^(\prime)\to(-4+3,5-7)=(-1,-2) \\ B^(\prime)\to(1+3,4-7)=(4,-3) \\ C^(\prime)\to(0+3,-3-7)=(3,-10) \end{gathered}`