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Question 508415: Three vertices (corners) of a square are: (-5, 4), (-5, -2), and (1,-2). Find the coordinates of the fourth vertex and then find the area of the square...
Describe the four directions a line can go and identify the name of the slope.
...
Fill in the blank:
a) Parallel lines have the ______________ slope.
b) Perpendicular lines have slopes that are ________________________ . . . more than one word goes in this blank....
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! Since all distances from vertex to vertex are equidistant (that is all the same length), then we can find the coordinate of the vertex using the distance formula.
recall that the vertex formula is sqrt((y2-y1)^2 + (x2-x1)^2)
Then sqrt((4+2)^2 + (-5 +5)^2) = sqrt(36) = 6
We know one side of the square, and since all sides are the same, we know that all distances are going to be length 6.
Let's look at something here.
(-5,4) ---> (-5,-2) travelled down
(-5,-2) ----> (1,-2) travelled right
(1,-2) ---> (-5,4) no apparent connection
(-5,4) top left
(-5,-2) bottom left
(1,-2) bottom right
(?, ?) top right
It seems like we need the top right corner.
Sorry, this would be much easier with a graph.
So we can either start at the top left or the bottom right.
Let's go with the bottom right.
In order to get to our mystery vertex, we need to travel up 6 units.
(1,-2+6) gives us (1,4) <---- Our answer
Let's try it from the top left and see if we get the same thing.
(-5+6, 4) = (1,4) <---- Same answer.
So our vertex is (1,4)
Now for the area. We already established that the side is 6. So if we take our formula A = s^2, then we easily find that the area is 36. <--- Answer
If this was not clear, please e-mail me at onlyyourmathtutor@yahoo.com.
Hope this helped!
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