Question

ph the image of square klmn after a translation 6 units left and 4 uni

Explanation:

Step1: Recall translation rule

For a translation $6$ units left and $4$ units up, the rule for a point $(x,y)$ is $(x - 6,y + 4)$.

Step2: Assume coordinates of square vertices

Let's assume the coordinates of the vertices of square $KLMN$ are $K(x_1,y_1)$, $L(x_2,y_2)$, $M(x_3,y_3)$, $N(x_4,y_4)$. From the graph, if we assume $K(4,- 5)$, $L(6,-5)$, $M(6,-3)$, $N(4,-3)$.

Step3: Apply translation rule to each vertex

For point $K(4,-5)$: $(4 - 6,-5 + 4)=(-2,-1)$.
For point $L(6,-5)$: $(6 - 6,-5 + 4)=(0,-1)$.
For point $M(6,-3)$: $(6 - 6,-3 + 4)=(0,1)$.
For point $N(4,-3)$: $(4 - 6,-3 + 4)=(-2,1)$.

Step4: Graph the new - square

Plot the points $(-2,-1)$, $(0,-1)$, $(0,1)$, $(-2,1)$ and connect them to form the translated square.

Answer:

Graph the square with vertices $(-2,-1)$, $(0,-1)$, $(0,1)$, $(-2,1)$ on the given coordinate - plane.