document.write( "Question 1045199: Two vertices of an equilateral triangle are (10,-4) and (0,6). Find the third vertex \n" ); document.write( "
Algebra.Com's Answer #660598 by Alan3354(69443)\"\" \"About 
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Two vertices of an equilateral triangle are (10,-4) and (0,6). Find the third vertex
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\n" ); document.write( "Label the points A(10,-4) and B(0,6)
\n" ); document.write( "Find the length of AB and the midpoint, label it M
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\n" ); document.write( "Length of AB = sqrt(diffy^2 + diffx^2) = sqrt(100+100) = 10sqrt(2)
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\n" ); document.write( "Find the average of x & y separately.
\n" ); document.write( "(10+0)/2 = 5
\n" ); document.write( "(-4+6)/2 = 1
\n" ); document.write( "M(5,1)
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\n" ); document.write( "Find the equation of the perpendicular bisector.
\n" ); document.write( "Slope of AB = 10/-10 = -1
\n" ); document.write( "Slope m of the perpendicular bisector = +1
\n" ); document.write( "--> y-1 = 1*(x-5)
\n" ); document.write( "y = x-4
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\n" ); document.write( "There are 2 vertices, C & D, both are on the line y = x-4
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\n" ); document.write( "The altitude of the triangle is 10.
\n" ); document.write( "Find the 2 points on y = x-4 10 units distance from M(5,1)
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\n" ); document.write( "d^2 = diffx^2 + diffy^2
\n" ); document.write( "100 = (x-5)^2 + (y-1)^2
\n" ); document.write( "Sub for y
\n" ); document.write( "100 = (x-5)^2 + (x-5)^2
\n" ); document.write( "(x-5)^2 = 50
\n" ); document.write( "x-5 = ħsqrt(50)
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\n" ); document.write( "x = 5 + sqrt(50), y = 1 + sqrt(50) --> Vertex C
\n" ); document.write( "x = 5 - sqrt(50), y = 1 - sqrt(50) --> Vertex D
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