Answer:
see attached
Explanation:
You want the given triangle reflected over the line y = -1/2x +1.
Reflection over a line
The transformation for reflection over the line y = mx +b is ...
(x, y) ⇒ (x(1-m²) +y(2m) -2mb, x(2m) +y(m²-1) +2b)/(1+m²)
Application
Using m=-1/2 and b=1, this transformation becomes ...
(x, y) ⇒ ((3x -4y +4)/5, (-4x -3y +8)/5)
Then for the vertex points, we get ...
(1, 3) ⇒ ((3·1 -4·3 +4)/5, (-4·1 -3·3 +8)/5) = (-5/5, -5/5) = (-1, -1)
(1, 8) ⇒ (3·1 -4·8 +4)/5, (-4·1 -3·8 +8)/5) = (-25/5, -20/5) = (-5, -4)
(6, 3) ⇒ (3·6 -4·3 +4)/5, (-4·6 -3·3 +8)/5) = (10/5, -25/5) = (2, -5)
Graph
The line of reflection is the perpendicular bisector of the segments connecting each point with its image. This fact can be used to find the reflection graphically.
The given line has slope -1/2, so a perpendicular line will have slope -1/(-1/2) = 2. A line with slope 2 can be drawn through each vertex point. The reflected point will be the same distance along that line on the other side of the line of reflection. The dashed lines in the attached figure show the idea.