document.write( "Question 87262: Can you please help with this question? Thanks Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5). \n" ); document.write( "
Algebra.Com's Answer #63199 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
First convert the equation into slope-intercept form\r
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Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)
Convert from standard form (Ax+By = C) to slope-intercept form (y = mx+b)


\"1x%2B3y=6\" Start with the given equation


\"1x%2B3y-1x=6-1x\" Subtract 1x from both sides


\"3y=-1x%2B6\" Simplify


\"%283y%29%2F%283%29=%28-1x%2B6%29%2F%283%29\" Divide both sides by 3 to isolate y


\"y+=+%28-1x%29%2F%283%29%2B%286%29%2F%283%29\" Break up the fraction on the right hand side


\"y+=+%28-1%2F3%29x%2B2\" Reduce and simplify


The original equation \"1x%2B3y=6\" (standard form) is equivalent to \"y+=+%28-1%2F3%29x%2B2\" (slope-intercept form)


The equation \"y+=+%28-1%2F3%29x%2B2\" is in the form \"y=mx%2Bb\" where \"m=-1%2F3\" is the slope and \"b=2\" is the y intercept.



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\n" ); document.write( "\n" ); document.write( "Now lets find the perpendicular line\r
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Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line

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\n" ); document.write( " Remember, any two perpendicular lines are negative reciprocals of each other. So if you're given the slope of \"-1%2F3\", you can find the perpendicular slope by this formula:
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\n" ); document.write( " \"m%5Bp%5D=-1%2Fm\" where \"m%5Bp%5D\" is the perpendicular slope
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\n" ); document.write( " \"m%5Bp%5D=-1%2F%28-1%2F3%29\" So plug in the given slope to find the perpendicular slope
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\n" ); document.write( " \"m%5Bp%5D=%28-1%2F1%29%283%2F-1%29\" When you divide fractions, you multiply the first fraction (which is really \"1%2F1\") by the reciprocal of the second
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\n" ); document.write( " \"m%5Bp%5D=3%2F1\" Multiply the fractions.
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\n" ); document.write( " So the perpendicular slope is \"3\"
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\n" ); document.write( " So now we know the slope of the unknown line is \"3\" (its the negative reciprocal of \"-1%2F3\" from the line \"y=%28-1%2F3%29%2Ax%2B2\").\n" ); document.write( "Also since the unknown line goes through (-3,5), we can find the equation by plugging in this info into the point-slope formula
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\n" ); document.write( " Point-Slope Formula:
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\n" ); document.write( " \"y-y%5B1%5D=m%28x-x%5B1%5D%29\" where m is the slope and (\"x%5B1%5D\",\"y%5B1%5D\") is the given point
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\n" ); document.write( " \"y-5=3%2A%28x%2B3%29\" Plug in \"m=3\", \"x%5B1%5D=-3\", and \"y%5B1%5D=5\"
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\n" ); document.write( " \"y-5=3%2Ax-%283%29%28-3%29\" Distribute \"3\"
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\n" ); document.write( " \"y-5=3%2Ax%2B9\" Multiply
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\n" ); document.write( " \"y=3%2Ax%2B9%2B5\"Add \"5\" to both sides to isolate y
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\n" ); document.write( " \"y=3%2Ax%2B14\" Combine like terms
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\n" ); document.write( " So the equation of the line that is perpendicular to \"y=%28-1%2F3%29%2Ax%2B2\" and goes through (\"-3\",\"5\") is \"y=3%2Ax%2B14\"
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\n" ); document.write( " So here are the graphs of the equations \"y=%28-1%2F3%29%2Ax%2B2\" and \"y=3%2Ax%2B14\"
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\n" ); document.write( " graph of the given equation \"y=%28-1%2F3%29%2Ax%2B2\" (red) and graph of the line \"y=3%2Ax%2B14\"(green) that is perpendicular to the given graph and goes through (\"-3\",\"5\")
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\n" ); document.write( "\n" ); document.write( "Now convert back to standard form\r
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Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)
Convert from slope-intercept form (y = mx+b) to standard form (Ax+By = C)


\"y+=+3x%2B14\" Start with the given equation


\"1y-3x+=+3x%2B14-3x\" Subtract 3x from both sides


\"-3x%2B1y+=+14\" Simplify


\"-1%2A%28-3x%2B1y%29+=+-1%2A%2814%29\" Multiply both sides by -1 to make the A coefficient positive (note: this step may be optional; it will depend on your teacher and/or textbook)


\"3x-1y+=+-14\" Distribute and simplify


The original equation \"y+=+3x%2B14\" (slope-intercept form) is equivalent to \"3x-1y+=+-14\" (standard form where A > 0)


The equation \"3x-1y+=+-14\" is in the form \"Ax%2BBy+=+C\" where \"A+=+3\", \"B+=+-1\" and \"C+=+-14\"


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