document.write( "Question 207074: How do you solve:
\n" ); document.write( "Point A (–4,1) is in the standard (x,y) coordinate plane. What must be the coordinates
\n" ); document.write( "of point B so that the line x = 2 is the perpendicular bisector of AB? \r
\n" ); document.write( "\n" ); document.write( "Thank you
\n" ); document.write( "

Algebra.Com's Answer #156556 by dlam5(15)\"\" \"About 
You can put this solution on YOUR website!
perp. bis. is also a bisector, meaning it must hit in the MIDDLE.\r
\n" ); document.write( "\n" ); document.write( "if A(-4,1) then call B(x,y).\r
\n" ); document.write( "\n" ); document.write( "if x=2, 2,1 is the midpt. of AB; as midpt. of something is a line; here, it is a horiz. straight line.\r
\n" ); document.write( "\n" ); document.write( "Now use midpt. formula.\r
\n" ); document.write( "\n" ); document.write( "-4 + x/2 , 1 +y/2 = 2,1\r
\n" ); document.write( "\n" ); document.write( "for x; set -4 + x/2=2; find x. x=8\r
\n" ); document.write( "\n" ); document.write( "for y; set 1+y/2 = 1; y=1\r
\n" ); document.write( "\n" ); document.write( "B(8,1); check.\r
\n" ); document.write( "\n" ); document.write( "is 2,1 between A and B? use midpt. to find out...\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "yes:)\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Edd
\n" ); document.write( "
\n" );