Question

1. reflection across y = -1

s(1, +5), n(1, +2), v(4, +2), a(2, +5)

Explanation:

Step1: Recall reflection formula

For a point $(x,y)$ reflected across the horizontal line $y = k$, the new - point is $(x,2k - y)$. Here $k=-1$.

Step2: Find the reflection of point $S(1,5)$

Substitute $x = 1,y = 5,k=-1$ into the formula. The new $y$ - coordinate is $2\times(-1)-5=-2 - 5=-7$, and the $x$ - coordinate remains the same. So the new point is $S'(1,-7)$.

Step3: Find the reflection of point $N(1,2)$

Substitute $x = 1,y = 2,k=-1$ into the formula. The new $y$ - coordinate is $2\times(-1)-2=-2 - 2=-4$, and the $x$ - coordinate remains the same. So the new point is $N'(1,-4)$.

Step4: Find the reflection of point $V(4,2)$

Substitute $x = 4,y = 2,k=-1$ into the formula. The new $y$ - coordinate is $2\times(-1)-2=-2 - 2=-4$, and the $x$ - coordinate remains the same. So the new point is $V'(4,-4)$.

Step5: Find the reflection of point $A(2,5)$

Substitute $x = 2,y = 5,k=-1$ into the formula. The new $y$ - coordinate is $2\times(-1)-5=-2 - 5=-7$, and the $x$ - coordinate remains the same. So the new point is $A'(2,-7)$.

Answer:

$S'(1,-7),N'(1,-4),V'(4,-4),A'(2,-7)$