To solve this problem, we need to reflect each vertex of the triangle [tex]\( NMO \)[/tex] over the vertical line [tex]\( x = -3 \)[/tex].
1. Reflecting Vertex [tex]\( N(-5, 2) \)[/tex] over [tex]\( x = -3 \)[/tex]:
- The distance from [tex]\( -5 \)[/tex] to [tex]\( -3 \)[/tex] is [tex]\( -3 - (-5) = 2 \)[/tex].
- To reflect it, we move [tex]\( 2 \)[/tex] units to the right of [tex]\( -3 \)[/tex]:
- New x-coordinate: [tex]\( -3 + 2 = -1 \)[/tex].
- The y-coordinate remains the same, which is [tex]\( 2 \)[/tex].
- So, the reflected vertex [tex]\( N' \)[/tex] is [tex]\( (-1, 2) \)[/tex].
2. Reflecting Vertex [tex]\( M(-2, 1) \)[/tex] over [tex]\( x = -3 \)[/tex]:
- The distance from [tex]\( -2 \)[/tex] to [tex]\( -3 \)[/tex] is [tex]\( -2 - (-3) = 1 \)[/tex].
- To reflect it, we move [tex]\( 1 \)[/tex] unit to the left of [tex]\( -3 \)[/tex]:
- New x-coordinate: [tex]\( -3 - 1 = -4 \)[/tex].
- The y-coordinate remains the same, which is [tex]\( 1 \)[/tex].
- So, the reflected vertex [tex]\( M' \)[/tex] is [tex]\( (-4, 1) \)[/tex].
3. Reflecting Vertex [tex]\( O(-3, 3) \)[/tex] over [tex]\( x = -3 \)[/tex]:
- Since [tex]\( O \)[/tex] is on the line [tex]\( x = -3 \)[/tex], the reflection will be the point itself.
- So, the reflected vertex [tex]\( O' \)[/tex] is [tex]\( (-3, 3) \)[/tex].
Therefore, the vertices of the image [tex]\( N'M'O' \)[/tex] after reflecting over the line [tex]\( x = -3 \)[/tex] are:
[tex]\[
N'(-1, 2), M'(-4, 1), O'(-3, 3)
\][/tex]
The correct choice from the given options is:
[tex]\[ N^{\prime}(-1,2), M^{\prime}(-4,1), O^{\prime}(-3,3) \][/tex]
