Question

Triangle XYZ has vertices X(-1, -2), Y(6, -3) , and Z(2, -5). Find the vertices of triangle X'Y'Z' after the translation of 2 units left and 1 unit up.

Answer 1

To obtain the coordinates of the vertices of triangle X'Y'Z' after the translation of 2 units left and 1 unit up, we simply subtract 2 from the x-coordinates of the vertices of triangle XYZ and also add 1 to the y-coordinates, as follows:

$\begin{gathered} X(-1,-2)\Longrightarrow X^1(-1-2,-2+1) \\ \text{thus:} \\ X^1(-3,-1) \end{gathered}$$\begin{gathered} Y(6,-3)\Longrightarrow Y^1(6-2,-3+1) \\ \text{thus}\colon \\ Y^1(4,-2) \end{gathered}$

and lastly:

$\begin{gathered} Z(2,-5)\Longrightarrow Y^1(2-2,-5+1) \\ \text{thus:} \\ Z^1(0,-4) \end{gathered}$

Therefore, the vertices of triangle X'Y'Z' after the translation of 2 units left and 1 unit up are:

X' (-3, -1)

Y' (4, -2)

Z' (0, -4)​