Given:
The triangle is ABC
Vertices of ABC is
[tex]\begin{gathered} A=(-4,3) \\ \\ B=(0,5) \\ \\ C=(-2,0) \end{gathered}[/tex]Find-:
The vertex after 4 units down
Explanation-:
The triangle is down, which means changing the coordinates of the y-axis
The y axis reduce by 4 units, then coordinates is
[tex]\begin{gathered} A=(-4,3) \\ \\ A\rightarrow A^{\prime} \\ \\ A^{\prime}=(-4,(3-4)) \\ \\ A^{\prime}=(-4,-1) \end{gathered}[/tex]The B' is
[tex]\begin{gathered} B=(0,5) \\ \\ B^{\prime}=(0,(5-4)) \\ \\ B^{\prime}=(0,1) \end{gathered}[/tex]The C' is
[tex]\begin{gathered} C^{\prime}=(-2,(0-4)) \\ \\ C^{\prime}=(-2,-4) \end{gathered}[/tex]So, the new coordinates are
[tex]\begin{gathered} A^{\prime}(-4,-1) \\ \\ B^{\prime}(0,1) \\ \\ C^{\prime}(-2,-4) \end{gathered}[/tex]