Question

the graph shows triangles ijk and ijk. which sequence of transformations maps ijk onto ijk? a rotation 90° counterclockwise around the origin followed by a translation right 1 unit and down 2 units a translation right 2 units and up 3 units followed by a reflection across the x - axis

Explanation:

Step1: Recall transformation rules

A 90 - degree counter - clockwise rotation around the origin changes a point $(x,y)$ to $(-y,x)$. A translation right $a$ units and down $b$ units changes a point $(x,y)$ to $(x + a,y - b)$. A translation right $m$ units and up $n$ units changes a point $(x,y)$ to $(x + m,y + n)$. A reflection across the $x$ - axis changes a point $(x,y)$ to $(x,-y)$.

Step2: Analyze the first option

For a 90 - degree counter - clockwise rotation around the origin, if $I=(-8,5)$, it becomes $I_1=(-5,-8)$. Then a translation right 1 unit and down 2 units makes it $I'=(-5 + 1,-8-2)=(-4,-10)$. This is incorrect.

Step3: Analyze the second option

Let's assume a point in $\triangle IJK$, say $I=(-8,5)$. A translation right 2 units and up 3 units makes it $I_1=(-8 + 2,5 + 3)=(-6,8)$. A reflection across the $x$ - axis makes it $I'=(-6,-8)$. For $J=(-6,2)$, a translation right 2 units and up 3 units makes it $J_1=(-6 + 2,2 + 3)=(-4,5)$. A reflection across the $x$ - axis makes it $J'=(-4,-5)$. For $K=(-3,5)$, a translation right 2 units and up 3 units makes it $K_1=(-3+2,5 + 3)=(-1,8)$. A reflection across the $x$ - axis makes it $K'=(-1,-8)$. This sequence of transformations maps $\triangle IJK$ onto $\triangle I'J'K'$.

Answer:

a translation right 2 units and up 3 units followed by a reflection across the x - axis