Question

the table shows the coordinates of the vertices of \\( \triangle jkl \\) and its image.

which translation maps \\( \triangle jkl \\) onto its image?
4 units down
4 units up
4 units to the right
4 units to the left

Explanation:

Step1: Analyze the change in y - coordinates

For point \(J(-3,0)\) and its image \(J'(-3,-4)\), the y - coordinate changes from \(y = 0\) to \(y=-4\). The change is \(0-(-4)= - 4\).
For point \(K(0,4)\) and its image \(K'(0,0)\), the y - coordinate changes from \(y = 4\) to \(y = 0\). The change is \(4 - 4=-4\).
For point \(L(2,3)\) and its image \(L'(2,-1)\), the y - coordinate changes from \(y = 3\) to \(y=-1\). The change is \(3-4=-4\).

Step2: Determine the translation rule

In a coordinate plane, if a point \((x,y)\) is translated to \((x,y + k)\), when \(k<0\), it is a vertical translation downwards. Here, since the change in the y - coordinate for each vertex is \(-4\) (i.e., \(y\to y-4\)), the translation is \(4\) units down.

Answer:

4 units down