document.write( "Question 37610: Find the two remaining vertices of square ABCD that has A=(-2,1), B=(6,-5) and a vertex in quadrant III. \n" ); document.write( "
Algebra.Com's Answer #23337 by venugopalramana(3286)\"\" \"About 
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Find the two remaining vertices of square ABCD that has A=(-2,1), B=(6,-5) and a vertex in quadrant III.
\n" ); document.write( "AB=SQRT((6+2)^2+(5+1)^2)=10
\n" ); document.write( "EQN. OF AB IS
\n" ); document.write( "(Y-1)=[(1+5)/(-6-2)](X+2)}
\n" ); document.write( "4y-4=-3x-6
\n" ); document.write( "3X+4Y=-2
\n" ); document.write( "ITS DISTANCE FROM ORIGIN IS 2/5
\n" ); document.write( "CD IS PARALLEL TO AB AND IS AT 10 DIUSTANCEB FROM AB OR 10+2/5 =52/5 DISTANCE FROM ORIGIN.
\n" ); document.write( "HENCE ITS EQN,IS
\n" ); document.write( "3X+4Y=-52................................I..OR...3H+4K=-52 AS C H,K LIES ON IT.
\n" ); document.write( "SLOPE OF AB =-3/4
\n" ); document.write( "SLOPE OF BC =4/3=
\n" ); document.write( "LET C BE H,K
\n" ); document.write( "SLOPE OF BC = (K+5)/(H-6)=4/3
\n" ); document.write( "3K+15=4H-24
\n" ); document.write( "4H-3K=39...........................II
\n" ); document.write( "SOLVING I AND II
\n" ); document.write( "H=0...K=-13.SO C IS (0,-13)
\n" ); document.write( "SIMILARLY FOR D WE CAN FIND \r
\n" ); document.write( "\n" ); document.write( "SLOPE OF AD =4/3=
\n" ); document.write( "LET D BE H,K
\n" ); document.write( "SLOPE OF AD = (K-1)/(H+2)=4/3
\n" ); document.write( "3K-3=4H+8
\n" ); document.write( "4H-3K=-11...........................III
\n" ); document.write( "SOLVING I AND III
\n" ); document.write( "H=-8 AND K=-7\r
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