document.write( "Question 89841: Find an equation of a line through (1,1) and (-2,0)in standard form. \n" ); document.write( "
Algebra.Com's Answer #65218 by jim_thompson5910(35256)\"\" \"About 
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Solved by pluggable solver: Finding the Equation of a Line
First lets find the slope through the points (\"1\",\"1\") and (\"-2\",\"0\")
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\n" ); document.write( " \"m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29\" Start with the slope formula (note: (\"x%5B1%5D\",\"y%5B1%5D\") is the first point (\"1\",\"1\") and (\"x%5B2%5D\",\"y%5B2%5D\") is the second point (\"-2\",\"0\"))
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\n" ); document.write( " \"m=%280-1%29%2F%28-2-1%29\" Plug in \"y%5B2%5D=0\",\"y%5B1%5D=1\",\"x%5B2%5D=-2\",\"x%5B1%5D=1\" (these are the coordinates of given points)
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\n" ); document.write( " \"m=+-1%2F-3\" Subtract the terms in the numerator \"0-1\" to get \"-1\". Subtract the terms in the denominator \"-2-1\" to get \"-3\"
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\n" ); document.write( " \"m=1%2F3\" Reduce
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\n" ); document.write( " So the slope is
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\n" ); document.write( " \"m=1%2F3\"
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\n" ); document.write( "Now let's use the point-slope formula to find the equation of the line:
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\n" ); document.write( " ------Point-Slope Formula------
\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 one of the given points
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\n" ); document.write( " So lets use the Point-Slope Formula to find the equation of the line
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\n" ); document.write( " \"y-1=%281%2F3%29%28x-1%29\" Plug in \"m=1%2F3\", \"x%5B1%5D=1\", and \"y%5B1%5D=1\" (these values are given)
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\n" ); document.write( " \"y-1=%281%2F3%29x%2B%281%2F3%29%28-1%29\" Distribute \"1%2F3\"
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\n" ); document.write( " \"y-1=%281%2F3%29x-1%2F3\" Multiply \"1%2F3\" and \"-1\" to get \"-1%2F3\"
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\n" ); document.write( " \"y=%281%2F3%29x-1%2F3%2B1\" Add \"1\" to both sides to isolate y
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\n" ); document.write( " \"y=%281%2F3%29x%2B2%2F3\" Combine like terms \"-1%2F3\" and \"1\" to get \"2%2F3\" (note: if you need help with combining fractions, check out this solver)
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\n" ); document.write( " Answer:
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\n" ); document.write( " So the equation of the line which goes through the points (\"1\",\"1\") and (\"-2\",\"0\") is:\"y=%281%2F3%29x%2B2%2F3\"
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\n" ); document.write( " The equation is now in \"y=mx%2Bb\" form (which is slope-intercept form) where the slope is \"m=1%2F3\" and the y-intercept is \"b=2%2F3\"
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\n" ); document.write( " Notice if we graph the equation \"y=%281%2F3%29x%2B2%2F3\" and plot the points (\"1\",\"1\") and (\"-2\",\"0\"), we get this: (note: if you need help with graphing, check out this solver)
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\n" ); document.write( " Graph of \"y=%281%2F3%29x%2B2%2F3\" through the points (\"1\",\"1\") and (\"-2\",\"0\")
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\n" ); document.write( " Notice how the two points lie on the line. This graphically verifies our answer.
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\n" ); document.write( "\n" ); document.write( "Now lets convert 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+=+%281%2F3%29x%2B2%2F3\" Start with the given equation


\"3%2Ay+=+3%2A%28%281%2F3%29x%2B2%2F3%29\" Multiply both sides by the LCD 3


\"3y+=+1x%2B2\" Distribute and multiply


\"3y-1x+=+1x%2B2-1x\" Subtract 1x from both sides


\"-1x%2B3y+=+2\" Simplify


\"-1%2A%28-1x%2B3y%29+=+-1%2A%282%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)


\"1x-3y+=+-2\" Distribute and simplify


The original equation \"y+=+%281%2F3%29x%2B2%2F3\" (slope-intercept form) is equivalent to \"1x-3y+=+-2\" (standard form where A > 0)


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


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