Question

reflect triangle jkl over the line y = -x, followed by a translation (x, y) → (x + 2, y + 1)
rotate triangle jkl 90° ccw about the origin, followed by a translation (x, y) → (x - 2, y + 1)
rotate triangle jkl 180° cw about the origin, followed by a translation (x, y) → (x - 2, y + 1)
translate triangle jkl, (x, y) → (x + 2, y + 1), followed by a rotation of 180° ccw about the origin.

Explanation:

Step1: Recall transformation rules

The rule for reflecting a point $(x,y)$ over the line $y = -x$ is $(x,y)\to(-y,-x)$. The rule for a $90^{\circ}$ counter - clockwise (CCW) rotation about the origin is $(x,y)\to(-y,x)$, for a $180^{\circ}$ clockwise (CW) or counter - clockwise rotation about the origin is $(x,y)\to(-x,-y)$. The translation rule $(x,y)\to(x + a,y + b)$ moves the point $a$ units to the right (if $a>0$) and $b$ units up (if $b>0$).

Step2: Analyze each option

• Option 1: Reflecting over $y=-x$ and then translating $(x,y)\to(x + 2,y + 1)$.
• Option 2: Rotating $90^{\circ}$ CCW about the origin gives $(x,y)\to(-y,x)$, then translating $(x,y)\to(x - 2,y + 1)$ gives $(-y-2,x + 1)$.
• Option 3: Rotating $180^{\circ}$ CW about the origin gives $(x,y)\to(-x,-y)$, then translating $(x,y)\to(x - 2,y + 1)$ gives $(-x-2,-y + 1)$.
• Option 4: Translating $(x,y)\to(x + 2,y + 1)$ first and then rotating $180^{\circ}$ CCW about the origin gives $(-(x + 2),-(y + 1))=(-x-2,-y - 1)$.

We can assume some coordinates for the vertices of $\triangle JKL$, say $J(x_1,y_1),K(x_2,y_2),L(x_3,y_3)$.
For a reflection over $y=-x$, $J(x_1,y_1)\to J'(-y_1,-x_1)$, then after translation $J'(-y_1,-x_1)\to J''(-y_1 + 2,-x_1+1)$.
By checking the orientation and position of the triangles visually and using the transformation rules, we find that the correct sequence is to reflect triangle $JKL$ over the line $y=-x$, followed by a translation $(x,y)\to(x + 2,y + 1)$.

Answer:

Reflect triangle JKL over the line y = -x, followed by a translation (x,y) → (x + 2,y + 1)