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)\"\" \"About 
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How do I find the slope throught (6,1) and perpendicular to y= 3/2x + 1/4
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\n" ); document.write( "Since the slope of:
\n" ); document.write( "y= 3/2x + 1/4
\n" ); document.write( "is 3/2
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\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
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\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
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