document.write( "Question 980907: A line that is perpendicular to the line segment connected by (4,-2) and (6,-8) but passes through the point (12,-5) wh
\n" ); document.write( "at is this is slope and General form \n" ); document.write( "

Algebra.Com's Answer #601969 by Cromlix(4381) $\"\"$ $\"About$

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Hi there,
\n" ); document.write( "First work out gradient:
\n" ); document.write( "Gradient = y2 - y1/x2 - x1
\n" ); document.write( "Utilising (4,-2) and (6, -8)
\n" ); document.write( "-8 -(-2)/6 - 4
\n" ); document.write( "-8 + 2/6 - 4
\n" ); document.write( "-6/2
\n" ); document.write( "= -3
\n" ); document.write( "If two lines are perpendicular to one another
\n" ); document.write( "then their gradients multiply together to give -1
\n" ); document.write( "m1 x m2 = -1
\n" ); document.write( "So, -3 x m2 = - 1
\n" ); document.write( "m2 = 1/3
\n" ); document.write( "Next use line equation:
\n" ); document.write( "y - b = m(x - a)
\n" ); document.write( "Utilising m = 1/3 and (12, -5)
\n" ); document.write( "y -(-5) = 1/3(x - 12)
\n" ); document.write( "y + 5 = 1/3(x - 12)
\n" ); document.write( "y + 5 = 1/3x - 4
\n" ); document.write( "y = 1/3x -4 - 5
\n" ); document.write( "y = 1/3x - 9
\n" ); document.write( "or
\n" ); document.write( "3y = x - 27
\n" ); document.write( "Hope this helps:-) \n" ); document.write( "

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