Question 988589
The equations of two sides of a square are y=3x-1 and x+3y-6=0. (0,-1) is one vertex of the square. Find the coordinates of the other vertices.
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(0,-1) is point A.
Find the intersection of the lines, that's a 2nd vertex, point B.
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y=3x-1
x+3y-6=0 --> y = (-1/3)x + 2
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x + 3(3x-1) = 6
10x = 9
x = 0.9, y = 1.7 --> B(0.9,1.7)
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The distance between the 2 points is the side length.
{{{s = sqrt(diffy^2 + diffx^2) = sqrt(2.7^2 + 0.9^2)}}}
{{{s = sqrt(8.1) = 9sqrt(10)/10}}}
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Find the line thru A(0,-1) parallel to x+3y-6=0
--> y = (-1/3)x - 1
A vertex (point D) is on this line distance s from point A.
The x-distance from A to D is 2.7, and the y-distance is 0.9
--> D (-2.7,-0.1)
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The side CD goes thru D with a slope of 3 (parallel to AB)
y + 0.1 = 3(x + 2.7)
y = 3x + 8
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Vertex C is the intersection of y = 3x + 8 and y = (-1/3)x + 2
3x + 8 = -x/3 + 2
10x/3 = -6
x = -1.8, y = 2.6
C(-1.8,2.6)