If you were to reflect this figure across the segment m, and then reflect it again across the perpendicular bisector of segment m, what would the coordinates of polygon JKSF be?
First, we need to reflect the polygon JKSF across the segment m. To do this, we need to find the distance from the points to the line and then make it an equal distance from the segment m. Although that is one way of reflecting, the equation of segment m is y=x, so we can just use the rule for reflecting across the line y=x, (x,y)-->(y,x). Using this rule, we can conclude that the points that are reflected across segment m are as follows:
Point J= (2,-1) --> (-1,2)
Point K=(2,-3)-->(-3, 2)
Point S=(3, -4)-->(-4,3)
Point F=(0,-4)-->(-4,0)
Once we reflect it, it should look like this:
First, we need to reflect the polygon JKSF across the segment m. To do this, we need to find the distance from the points to the line and then make it an equal distance from the segment m. Although that is one way of reflecting, the equation of segment m is y=x, so we can just use the rule for reflecting across the line y=x, (x,y)-->(y,x). Using this rule, we can conclude that the points that are reflected across segment m are as follows:
Point J= (2,-1) --> (-1,2)
Point K=(2,-3)-->(-3, 2)
Point S=(3, -4)-->(-4,3)
Point F=(0,-4)-->(-4,0)
Once we reflect it, it should look like this:
Next, we need to find the perpendicular bisector of the segment m. To start, we need to find the equation of segment m, which is y=x. In order to find the perpendicular bisector of m, you need to find the opposite reciprocal slope. The opposite reciprocal of 1 is -1. Next, we need to find the midpoint of segment m. To do this, you can use the midpoint formula. Plug the numbers into the formula, and you would get (-4+4/2, -4+4/2), which would then simplify down to (0/2, 0/2), and would finally simplify to (0,0) which is also the origin. Now that we know the midpoint and the slope of the line, we can put the equation together to make the equation y=-x.
The next step is to reflect the the figure J'K'S'F' across the line y=-1. The mapping rule for the reflection y=-x is
(x,y)-->(-y,-x). If we plug in the points to the equation, we get the answer to the question:
Point J=(-1,2)-->(-2,1)
Point K=(-3,2)-->(-2,3)
Point S=(-4,3)-->(-3,4)
Point F= (-4,0)-->(0,4)
(x,y)-->(-y,-x). If we plug in the points to the equation, we get the answer to the question:
Point J=(-1,2)-->(-2,1)
Point K=(-3,2)-->(-2,3)
Point S=(-4,3)-->(-3,4)
Point F= (-4,0)-->(0,4)