Question

△abc is reflected about the line ( y = -x ) to give ( \triangle abc ) with vertices ( a(-1, 1) ), ( b(-2, -1) ), ( c(-1, 0) ). what are the vertices of ( \triangle abc )?
a. ( a(1, -1) ), ( b(-1, -2) ), ( c(0, -1) )
b. ( a(-1, 1) ), ( b(1, 2) ), ( c(0, 1) )
c. ( a(-1, -1) ), ( b(-2, -1) ), ( c(-1, 0) )
d. ( a(1, 1) ), ( b(2, -1) ), ( c(1, 0) )
e. ( a(1, 2) ), ( b(-1, 1) ), ( c(0, 1) )

Explanation:

Step1: Recall reflection inverse rule

The reflection of a point $(x,y)$ over $y=-x$ is $(-y,-x)$. To reverse this, if the reflected point is $(x',y')$, then the original point $(x,y)$ satisfies $x'=-y$ and $y'=-x$, so $y=-x'$ and $x=-y'$. Thus, original point is $(-y', -x')$.

Step2: Find vertex A

Take $A'(-1,1)$. Compute original $A$: $(-1, 1) \to (-1, -(-1)) = (-1,1)$? No, apply $(-y', -x')$: $y'=1, x'=-1$, so $(-1, -(-1)) = (-1,1)$? Wait no, correct: if $(x',y') = (-y,-x)$, then solving for $x,y$: $x = -y'$, $y = -x'$. So for $A'(-1,1)$: $x=-1$, $y=-(-1)=1$? No, wait, let's do it properly:
If $(x', y')$ is reflection of $(x,y)$ over $y=-x$, then $x' = -y$, $y' = -x$. So solving for $x$ and $y$: $y = -x'$, $x = -y'$.
So for $A'(-1, 1)$: $x = -1$, $y = -(-1) = 1$? No, $x = -y' = -1$, $y = -x' = -(-1) = 1$ → $A(-1,1)$

Step3: Find vertex B

Take $B'(-2,-1)$. $x = -y' = -(-1)=1$, $y=-x' = -(-2)=2$ → $B(1,2)$

Step4: Find vertex C

Take $C'(-1,0)$. $x=-y' = -0=0$, $y=-x' = -(-1)=1$ → $C(0,1)$

Answer:

B. $A(-1, 1), B(1, 2), C(0, 1)$