We can start by plotting the triangle JAG and the line y = 2 to see how we can solve it:
We can see the line y = 2 (red line) over which we have to reflect the point J, A and G.
To do that we have to look at the distance of each point to the line.
This distance will be the distance at which the image point will be located, opposite to the the original point.
As y = 2 is an horizontal line, the distance is vertical and the x-coordinate of the original points and the images will remain equal. Only the y-coordinate will change.
We can sketch the procedure as:
Then, for example, poing G is 1 unit below y = 2, so its image G' will be one unit over the line y = 2, locating at (2,3).
Point J is 2 units above y = 2, so J' will be 2 units below y = 2.
Point A is 3 units above y = 2, so A' will be 3 untis below y = 2
We can plot the remaining points as:
The red triangle is the original triangle and the blue triangle is the image.
Answer:
The coordinates are J'=(1,0), A'=(3,-1) and G'=(2,3).
