Question

After a translation 5 units left and 6 units down.

Answer 1

We will investigate the translation transformation applied on a figure defined on a cartesian coordinate grid.

Translation transformation deals with the displacement of the primary vertices of a figure like a square JKLM with certain number off units in horizontal and/or vertical direction.

We will investigate the type of translations that are possible as follows:

`(\text{ x , y ) }\to\text{ ( x + a , y + b )}`

Where,

`\begin{gathered} a\colon\text{ Number of horizontal units shifts} \\ b\text{ : Number of vertical units shifts} \end{gathered}`

AND,

`\begin{gathered} a\text{ > 0 }\ldots\text{ Right shift} \\ a\text{ < 0 }\ldots\text{ Left shift} \\ a\text{ = 0 }\ldots\text{ No horizontal shift} \\ \\ b\text{ > 0 }\ldots\text{ Up shift} \\ b\text{ < 0 }\ldots\text{ Down shift} \\ b\text{ = 0 }\ldots\text{ No Vertical shift} \end{gathered}`

We will investigate how to use the above rule to determine the image of the square JKLM.

We will translate each vertex 5 units left and 6 units down! We will assign the constants a value according to the translation units:

`\begin{gathered} a\text{ = -5} \\ b\text{ = -6} \end{gathered}`

The general rule that will be applied to the vertices of the square JKLM:

`(\text{ x , y ) }\to\text{ ( x - 5 , y - 6 )}`

The four coordinate of the square are:

`J\text{ ( 1 , -3 ) , K ( 9 , -3 ) , L ( 9 , 5 ) , M ( 1 , 5 )}`

Applying the general rule to determine the image of all vertices:

`\begin{gathered} J\text{ ( 1 , -3 ) }\to\text{ J' ( 1 -5 , -3 - 6 ) }\to\text{ J' ( -4 , -9 )} \\ K\text{ ( 9 , -3 ) }\to\text{ K' ( 9 -5 , -3 - 6 ) }\to\text{ K' ( 4 , -9 )} \\ L\text{ ( 9 , 5 ) }\to\text{ L' ( 9 -5 , 5 - 6 ) }\to\text{ L' ( 4 , -1 )} \\ M\text{ ( 1 , 5 ) }\to\text{ M' ( 1 - 5 , 5 - 6 ) }\to\text{ M' ( -4 , -1 ) } \end{gathered}`

The image of the square JKLM is as follows:

`J^(\prime)\text{ ( -4 , -9 ) , K' ( 4 , -9 ) , L' ( 4 , -1 ) , M' ( -4 , -1 )}`

We can plot these images on the grid as follows: