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2 votes
Need help on number 8

Need help on number 8-example-1
User Roxerio
by
4.3k points

2 Answers

4 votes

Answer:

F(y) = 3y + 5x+5x this keeps F perpendicular.

F(x) = 10x +3y

When E stays constant (0,0) and F stays constant (0,10)

when f = 10

Then the lines are perpendicular

We can rephrase this as When EFF = (0,0) (0,10) (5,3) changes to EFF (0,0) (0,10) (10,3) then FF becomes perpendicular with FE.

Step-by-step explanation:

User Bajaj
by
4.8k points
1 vote

Answer for the first box: y = 0

Answer for the second box: y = x+1

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Step-by-step explanation:

The line y = 0 is the horizontal line that overlaps the x axis perfectly. If the line of reflection is y = 0, then you reflect over the x axis to have F = (6,3) move to F' = (6,-3). The perpendicular bisector of FF' is y = 0, which is exactly the line of reflection. Point E is on the line y = 0.

Now create any line you want that does not go through point E. One such line is y = x+1. This line of reflection will play the role as the perpendicular bisector of FF'. Point E is not on the line because plugging x = 0 does not lead to y = 0, and instead it leads to y = 1.

So as you can see, Kari's claim is not 100% correct as there are counter-examples to show it doesn't always work.

User MisterMetaphor
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4.2k points