document.write( "Question 365049: How do I find the slope throught (6,1) and perpendicular to y= 3/2x + 1/4 \n" ); document.write( "
Algebra.Com's Answer #260283 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! How do I find the slope throught (6,1) and perpendicular to y= 3/2x + 1/4 \n" ); document.write( ". \n" ); document.write( "Since the slope of: \n" ); document.write( "y= 3/2x + 1/4 \n" ); document.write( "is 3/2 \n" ); document.write( ". \n" ); document.write( "a line perpendicular to it must be the \"negative reciprocal\". \n" ); document.write( "That is, our new slope multiplied by 3/2 must be -1: \n" ); document.write( "(3/2)m = -1 \n" ); document.write( "3m = -2 \n" ); document.write( "m = -2/3 \n" ); document.write( ". \n" ); document.write( "Now, if you want the equation of the line passing through (6,1) with a slope of -2/3 plug it into the \"point-slope\" form: \n" ); document.write( "y - y1 = m(x - x1) \n" ); document.write( "y - 1 = (-2/3)(x - 6) \n" ); document.write( "y - 1 = (-2/3)x + 4 \n" ); document.write( "y = (-2/3)x + 5 (this is what they're looking for in \"slope-intercept\" form)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |