Question

the graph shows triangles klm and klm. which sequence of transformations maps klm onto klm? a rotation 90° clockwise around the origin followed by a translation right 1 unit and up 1 unit a rotation 180° around the origin followed by a translation right 9 units and down 1 unit a reflection across the y - axis followed by a translation right 11 units and down 10 units

Explanation:

Step1: Recall rotation rules

A 90 - degree clock - wise rotation around the origin has the rule $(x,y)\to(y, - x)$. A 180 - degree rotation around the origin has the rule $(x,y)\to(-x,-y)$. A reflection across the y - axis has the rule $(x,y)\to(-x,y)$.

Step2: Consider rotation 90 - degree clockwise

Let's assume a point $(x,y)$ in $\triangle KLM$. After a 90 - degree clockwise rotation around the origin, it becomes $(y, - x)$. Then, a translation right 1 unit and up 1 unit gives $(y + 1,-x + 1)$. This does not match the transformation from $\triangle KLM$ to $\triangle K'L'M'$.

Step3: Consider rotation 180 - degree

Let a point in $\triangle KLM$ be $(x,y)$. After a 180 - degree rotation around the origin, it is $(-x,-y)$. Then, a translation right 9 units and down 1 unit gives $(-x + 9,-y-1)$. This does not match the transformation from $\triangle KLM$ to $\triangle K'L'M'$.

Step4: Consider reflection across y - axis

Let a point $(x,y)$ in $\triangle KLM$. After reflection across the y - axis, it is $(-x,y)$. Then a translation right 11 units and down 10 units gives $(-x + 11,y - 10)$.
Let's take a vertex of $\triangle KLM$, say $K(7,8)$. After reflection across the y - axis, $K$ becomes $(-7,8)$. After translation right 11 units and down 10 units, $(-7+11,8 - 10)=(4,-2)$ which is the approximate location of $K'$.

Answer:

a reflection across the y - axis followed by a translation right 11 units and down 10 units