SOLUTION: Show that the points (12,9),(20,-6),(5,-14), and (-3,1) are the vertices of a square. What is the length of the diagonal?

Question 1123420: Show that the points (12,9),(20,-6),(5,-14), and (-3,1) are the vertices of a square. What is the length of the diagonal?

Found 2 solutions by Boreal, Alan3354:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Look at the distances of each of these from each other using the distance formula d=sqrt(chg x^2+chg y^2)
sqrt (289) for the first pair and for the adjacent pairs so sides are equal.
The pairs should have orthogonal or perpendicular slopes
the first pair -15/8
the second pair -8/-15 or 8/15, so they are perpendicular
and the other pairs do likewise.
Therefore, the diagonals are (12, 9) and (5, -14) and because it is a square, they should be sqrt(2)*s in length or sqrt(578)
The distance between those two points is sqrt (49+23^2) or sqrt (578)
Doing (20, -6) and (-3, 1) gives the same sqrt (578) distance.
These points are the vertices of a square.
The sides are equal
The sides are perpendicular
The diagonals are s*sqrt(2) more than the sides.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Show that the points (12,9),(20,-6),(5,-14), and (-3,1) are the vertices of a square. What is the length of the diagonal?
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Label the points.
A(12,9), B(20,-6), C(5,-14), and D(-3,1)
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Find the lengths of AB, BC, CD and DA.
If they're equal, it's at least a parallelogram, and the points are labelled in order around it.
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If, in addition, AB is perpendicular to BC, it's a square.
Diagonal = AB*sqrt(2)

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